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m  MEMORIAL 
Irving  Strlngham 


J 


-^ 


A 

PRACTICAL    APPLICATION 

OF 

THE   PRINCIPLES  OF    GEOMETRY 

TO  THB 

MENSURATION 

or 

SUPERFICIES   AND   SOLIDS. 


▲DAmB  TO 


THE  METHOD  OP  INSTRUCTION  IN  SCHOOLS  AND  ACADEMXEB. 


BY  JEREMIAH  DAY,  D.D.  LL.D. 

I.4TS  PUMOBBT  Or  TAU  OOLLBVC 


NEW   YORK: 

PUBLISHED  BY  MARK  H.   NEWMAN  <k  CO., 

No.   199  BROADWAY. 

1848. 


Entered,  according  to  Act  of  Congress,  in  the  year  1848,  by 

JEREMIAH    DAY, 

In  the  Clerk's  OfBce  of  the  District  Court  of  the  Unit«d  States  for  the 
Southern  District  of  New  York. 


THOMAS    B.    SMITH,    STERKOTTPER,  J.    D.    BEDFORD,   PRINTER, 

216  WILLIAM  STREET,  K.  Y.  138   FULTON    STREET. 


CONTENTS, 


Sectioh    L  Areas  of  figures  bonnded  by  right  lines,      ...  5 

II.  The  Quadrature  of  the  Circle  and  its  parts,  .    .  19 

Promiscnoos  examples  of  Areas, 34 

m.  Solids  bounded  by  plane  surfaces, 37 

rV.  The  Cylinder,  Cone,  and  Sphere, 66 

Promiscuous  examples  of  Solids, 76 

V.  Isoperimetry, 78 

APPENDIX 

Gauging  of  Casks, 92 

Notes,                      99 


SECTION  I. 


AREAS    OF    FIGURES    BOUNDED   BT    RIGHT    LINES. 


Art.  1.  The  following  definitions,  which  are  nearly  the 
same  as  in  Euclid,  are  inserted  here  for  the  convenience  of 
reference. 

I.  Four-sided  figures  have  different  names,  according  to 
the  relative  position  and  length  of  the  sides.  A  parallelo- 
gram has  its  opposite  sides  equal  and  parallel,  as  ABCD. 

1 


D| 1 r ^ |C 

a' ' ' ' 1 


(Fig.  2.)  A  rectangle,  or  right  parallelogram,  has  its  oppo- 
site sides  equal,  and  all  its  angles  right  angles  ;  as  AC. 
(Fig.  1.)  A  square  has  all  its  sides  equal,  and  all  its  angles 
right  angles ;  as  ABGH.  (Fig.  3.)    A  rhxmibus  has  all  its 


sides  equal,  and  its  anL^ 
rJwmJboid  has  its  opposi  \ 


e ;  as  ABCD.  (Fig.  3.)     A 
[ual,  and  its  angles  obUque  \^ 


as  ABCD.  (Fig.  2.)  A  trapezoid  has  only  two  of  its  sides 
parallel ;  as  ABCD.  (Fig.  4.)  Any  other  four  sided  figilre 
is  called  a  trapezium. 


6  MENSURATION    OF   PLANE   SUHFACES. 

II.  A  figure  which  has  more  than  four  sides  is  called  a 
polygon.  A  regular  polygon  has  all  its  sides  equal,  and  all 
its  angles  equal. 

in.  The  height  of  a  triangle 
is  the  length  of  a  perpen- 
dicular, drawn  from  one  of  the 
angles  to  the  opposite  side ;  as 
CP.  The  height  of  2kfour  sided 
figure  is  the  perpendicular  dis- 
tance between  two  of  its  par- 
allel sides ;  as  CP.  (Fig.  4.) 

IV.  The  area  or  superficial  contents  of  a  figure  is  the 
space  contained  within  the  line  or  lines  by  which  the  figure 
is  bounded. 

2.  In  calculating  areas,  some  particular  portion  of  surface 
is  fixed  upon,  as  the  measuring  unit,  with  which  the  given 
figure  is  to  be  compared.  This  is  commonly  a  square  ;  as  a 
square  inch,  a  square  foot,  a  square  rod,  &c.  For  this  rea- 
son, determining  the  quantity  of  surface  in  a  figure  is  called 
squaring  it,  or  finding  its  quadrature ;  that  is,  finding  a 
square  or  number  of  squares  to  which  it  is  equal. 

3.  The  superficial  unit  has  generally  the  same  name,  as 
the  linear  unit  which  forms  the  side  of  the  square. 

The  side  of  a  square  inch  is  a  linear  inch ; 
of  a  square  foot,  a  linear  foot ; 
of  a  square  rod,  a  linear  rod,  &;c. 
There  are  some  superficial  measures,  however,  which  have 
no  corresponding  denominations  of  length.     The  acre,  for 
instance,  is  not  a  square  which  has  a  line  of  the  same  name 
for  its  side. 

The  following  tables  contain  the  linear  measures  in  com- 
mon use,  with  their  corresponding  square  measures. 


MENSURATION   OF   PLANE   SURPAOBS. 


Linear  Measures. 

Square  Measures. 

12     inches  =1  foot. 

144     inches  =1  foot. 

3     feet       =1  yard. 

9     feet      =1  yard. 

6     feet       =1  fathom. 

36     feet       =1  fathom. 

16i  feet       =1  rod. 

272i  feet       =1  rod. 

5i  yards    =1  rod. 

30-I-  yards    =1  rod. 

4     rods      =1  chain. 

16     rods      =1  chain. 

40     rods      =1  furlong. 

1600     rods      =1  furlong. 

320     rods      =1  mile. 


102400"    rods      =1  mile. 


An  acre  contains  160  square  rods,  or  10  square  chains. 
By  reducing  the  denominations  of  square  measure,  it  will 
be  seen  that 

1  sq.  mi!e=610  acre8=102400roda=2787»100  feet  =4014489600  inchca. 
1  acre=iachain8=160  rods=:43560  feet=:6272640  inches. 

The  fundamental  problem  in  the  mensuration  of  super- 
ficies is  the  very  simple  one  of  determining  the  area  of  a 
riffht  parallelogram.  The  contents  of  other  figures,  partic- 
ularly those  which  are  rectilinear,  may  be  obtained  by  find- 
ing parallelograms  which  are  equal  to  them,  according  to  the 
principles  laid  down  in  Euclid. 

Problem  I. 

To  find  the  area  of  a  parallelogram,  square,  rhombus,  or 
rliomboid. 


4.   .NL  .  LENGTH  BY  Tr?r:   T.V 

or  BREAD! 

It  is  b  evident  that  the  number  of 
square  inches  in  the  parallelogram  AC 
is  equal  to  the  number  of  linear 
inches  in  the  length  AB,  repeated  as 
many  times  as  there  are  inches  in  the 
breadth  BC.  For  more  particular 
illustration  of  this  see  AJg.  886 — 389. 


••"'VDICULAR    llEIGIir 


I1~M.__ 


8 


MENSURATION    OF   PLANE   SURFACES. 


The  oblique  parallelogram  or  rhomboid  ABCD,  (Fig.  2.) 
is  equal  to  the  right  parallelogram  GHCD.  (Euc.  36. 1.)*  The 


area,  therefore,  is  equal  to  the  length  AB  multiplied  into  the 
perpendicular  height  HC.  And  the  rhombus  ABCD,  (Fig.  3.) 
is  equal  to  ihe  parallelogram  ABGH.  As  the  sides  of  a 
square  are  all  equal,  its  area  is  found,  by  multiplying  one  of 
the  sides  into  itself. 

Ex.  1.  How  many  square  feet  are  there  in  a  floor  23|-  feet 
long,  and  18  feet  broad  ?  Ans.  23^X18=423. 

2.  What  are  the  contents  of  a  piece  of  ground  which  is 
66  feet  square  ?  Ans.  4356  sq.  feet=16  sq.  rods. 

3.  How  many  square  feet  are  there  in  the  four  sides  of  a 
room  which  is  22  feet  long,  1*7  feet  broad,  and  11  feet  high  ? 

Ans.  858. 

Art.  5.  If  the  sides  and  angles  of  a  parallelogram  are 
given,  the  perpendicular  height  may  be  easily  found  by  trig- 
onometry. Thus,  CH  (Fig.  2.)  is  the  perpendicular  of  a 
right  angled  triangle,  of  which  BC  is  the  hypothenuse. 
Then,  (Trig,  134.) 

R  :  BC  :  :  sin  B  :  CH. 

The  area  is  obtained  by  multiplying  CH  thus  found,  into 
the  length  AB. 


*  Thomson's  Legendre,  1.  5. 


MENSURATION  OF  PLANE  SURFACES. 

Or,  to  reduce  the  two  operations  to  one, 

As  radius, 

To  the  sine  of  any  angle  of  a  parallelogram  ; 

So  is  the  product  of  the  sides  including  that  angle, 

To  the  area  of  the  parallelogram. 

For  the  arm=ABxCH,  (Fig.  2.)  But  CH=^^^^'"  ^ 


R 
Therefore, 

Thearea^^^^^^^'''''  ^. Or,R : sinB : : ABxBC : thearea. 
R 

Ex.  If  the  side  AB  be  58  rods,  BC  42  rods,  and  the  angle 
B  63°,  what  is  the  area  of  the  parallelogram  ? 

As  radius  10.00000 

To  the  sine  of  B  63°               9.94988 

(  So  is  the  product  of  AB  58                 1.76343 

\  Into  BC  (Trig.  39.)  42                 1.62325 

To  the  area  2170.5  sq.  rods  3.33656 

2.  If  the  side  of  a  rhombus  is  67  feet,  and  one  of  the 
angles  73**,  what  is  the  area  ?  Ans.  4292.7  feet. 

C.  When  the  dimensions  are  given  in  feet  and  inches,  the 
multiplication  may  be  conveniently  performed  by  the  arith- 
metical rule  of  Duodecimals  ;  in  which  each  inferior  denom- 
ination is  one-twelfth  of  the  next  higher.  Considering  a  foot 
as  the  measuring  unit,  a  prime  is  the  twelfth  part  of  a  foot ; 
a  second,  the  twelfth  part  of  a  prime,  kc.  It  is  to  be  ob- 
served, that,  in  measures  of  length,  inches  are  primes  ;  but  in 
superficial  measure  they  are  seconds.  In  both,  a  prime  is  -^ 
of  a  foot.  But  iV  of  a  square  foot  is  a  parallelogram,  a  foot 
long  and  an  inch  broad.  The  twelfth  part  of  this  is  a  square 
inch,  w^hich  is  i -J  y  of  a  square  foot. 


10  MENSURATION  OF  PLANE  SURFACES. 

Ex.  1.  What  is  the  surface  of  aboard  9  feet  5  inches,  by 
2  feet  1  inches. 

F 

9     5' 

2     '7 


18   10 

5     5     11 
24     3     11^  or  24  feet  47  inches. 

2.  How  many  feet  of  glass  are  there  in  a  window  4  feet 
11  inches  high,  and  3  feet  5  inches  broad  ? 

Ans.  16  F.  9'  1",  or  16  feet  115  inches. 

Y.  If  the  area  and  one  side  of  a  parallelogram  be  given, 
the  other  side  may  be  found  by  dividing  the  area  hy  the 
given  side.  And  if  the  area  of  a  square  be  given,  the  side 
may  be  found  by  extracting  the  square  root  of  the  area.  This 
is  merely  reversing  the  rule  in  Art.  4.     See  Alg.  520,  521. 

Ex.  1 .  What  is  the  breadth  of  a  piece  of  cloth  which  is 
36  yds.  long,  and  which  contains  63  square  yards. 

Ans.  If  yds. 

2.  What  is  the  side  of  a  square  piece  of  land  containing 
289  square  rods  ? 

3.  How  many  yards  of  carpeting  1^  yard  wide,  will  cover 
a  floor  30  feet  long  and  22^-  broad  ? 

Ans.  30X22ifeet=10xH='75  yds.  And '75-Mi=60. 

4.  What  is  the  side  of  a  square  which  is  equal  to  a  par- 
allelogram 936  feet  long  and  104  broad  ? 

5.  How  many  panes  of  8  by  10  glass  are  there,  in  a  win- 
dow 5  feet  high,  and  2  feet  8  inches  broad  ? 


MENSURATION    OF    PLANE    SCRFACES. 


11 


Problem  II. 

To  find  the  area  of  a  triangle. 

8.  Rule  L  Multiply  onb  side  by  half  the  perpen- 
dicular from  the  opposite  angle.  Or,  multiply  half  the 
side  by  the  perpendicular,  Or,  multiply  the  whole  side  by 
the  perpendicular  and  take  half  the  product. 

The  area  of  the  triangle  ABC, 
b  equal  to  ^  PC  X  AB,  because 
a  parallelogram  of  the  same 
base  and  height  is  equal  to  PC 
X  AB,  (Art.  4.)  and  by  Euc.  41, 
1,*  the  triangle  is  half  the  par- 
allelogram. 

Ex.  1.  If  AB  be  65  feet,  and  PC  31.2,  what  is  the  area 
of  the  triangle?  Ans.   1014  square  feet. 

2.  What  is  the  surface  of  a  triangular  board,  whose  base 
is  3  feet  2  inches,  and  perpendicular  height  2  feet  9  inches  ? 
Ans.  4  F.  4'  3",  or  4  feet  51  inches. 


9.  If  two  sides  of  a  triangle  and  the  included  angle,  are 
given,  the  perpendicular  on  one  of  these  sides  may  be  easily 
found  by  rectangular  trigonometry.  And  the  area  may  be 
calculated  in  the  same  man- 
ner as  the  area  of  a  parallel- 
og^m  in  Art.  5.  In  the  tri- 
angle ABC, 

R  :  BC  : :  sin  B  :  CH 


And  because  the  triangle  is  half  the  parallelogram  of  the 
same  base  and  height, 


*  Thomson's  Legendre,  2.  4. 


12 


MENSURATION  OF  PLANE  SURFACES. 


As  radius, 

To  the  sine  of  any  angle  of  a  triangle  ; 

So  is  the  product  of  the  sides  including  that  angle, 

To  twice  the  area  of  the  triangle.  (Art.  5.) 

Ex.  If  AC  be  39  feet,  AB 
65  feet,  and  the  angle  at  A  53° 
7'  48",  what  is  the  area  of  the 
triangle  ?  Ans.  1014  square  feet. 

9.  &.  If  one  side  and  the  angles 
are  given  ;  then 

As  the  product  of  radius  and  the  sine  of  the  angle  oppo- 
site the  given  side, 

To  the  product  of  the  sines  of  the  two  other  angles; 
So  is  the  square  of  the  given  side. 
To  twice  the  area  of  the  triangle. 

If  PC  be  perpendicular  to  AB. 

R  :  sin  B  :  :  BC   :  CP 
sin  ACB  :  sin  A  :  :  AB  :  BC 

Therefore,  (Alg.  351,  345.) 
R  X  sin  ACB  :  sin  A  X  sin  B  : :  AB  X  BC  :  CP  X  BC  : : 
AB^  :  ABxCP— twice  the  area  of  the  triangle. 

Ex.  If  one  side  of  a  triangle  be  57  feet,  and  the  angles  at 
the  ends  of  this  side  50°  and  60°,  what  is  the  area? 

Ans.  1147  sq.  feet. 

10.  If  the  sides  only  of  a  triangle  are  given,  an  angle  may 
be  found,  by  oblique  trigonometry,  Case  IV,  and  then  the 
perpendicular  and  the  area  may  be  calculated.  But  the  area 
may  be  more  directly  obtained,  by  the  following  method. 

Rule  II.  When  the  three  sides  are  given,  from  half  their 
sum  subtract  each  side  severally,  multiply  together  the  lialf 
sum  and  the  three  femainders,  and  extract  the  square  root  of 
the  product. 


MENSURATION  Of  PLANE  SURFACES. 


13 


If  the  sides  of  the  triangle  are  a,  h,  and  c,  andif  A=half 
their  sum,  then 


The  area=Vhx{hr—a)x{fir-h)X{h—c) 

Ex.  1.  In  the  triangle  ABC, 
given  the  sides  a  52  feet,  6  39, 
and  c  65  ;  to  find  the  side  of  a 
square  which  has  the  same  area 

as  tlio  Irianrjle. 


^— a=26 


A— 6=39 
A— c=13 


Then  the  area=v78x  26X39X13= 1014  square  feet. 
By  logarithms. 


The  half  sum 

=  78 

1.89209 

First  remainder 

=  20 

1.41497 

Second     do. 

=39 

1.59106 

Third       do. 

=  13 

1.11394 

2)6.01206 
2)3.00603 


The  area  required^        =1014 

Side  of  the  square      =31.843  (Trig.  47.)     1.50301 

2.  If  the  sides  of  a  triangle  are  134,  108,  and  80  rods, 
what  is  the  area?  Ans.  4319. 

3.  What  is  the  area  of  a  triangle  whose  sides  are  371, 264, 
and  225  feet  ? 


Problem  III. 
To  find  the  area  of  a  trapezoid. 
21.    Multiply  half  the   sum   op  the    parallel   sides 

INTO    their   perpendicular   DISTANCE. 


14 


MENSURATION  OF  PLANE  SURFACES. 


The  area  of  the  trapezoid 
ABCD,  is  equal  to  half  the 
sum  of  the  sides  AB  and  CD, 
multipHed  into  the  perpendic- 
ular distance  PC  or  AH.  For 
the  whole  figure  is  made  up  of 
the  two  triangles  ABC  and 
ADC  ;  the  area  of  the  first  of  which  is  equal  to  the  product 
of  half  the  base  AB  into  the  perpendicular  PC,  (Art.  8.) 
and  the  area  of  the  other  is  equal  to  the  product  of  half  the 
base  DC  into  the  perpendicular  AH  or  PC. 

Ex.  If  AB  be  46   feet,  BC   31,  DC  38,  and  the  angle  B 
•70°,  what  is  the  area  of  the  trapezoid  ? 

R  :  BC  :  :  sin  B  :  PC  =  29.13.    And 42x29. 13  =  1223^. 

2.  What  are  the  contents  of  a  field  which  has  two  par- 
allel sides  Q5  and  38  rods,  distant  from  each  other  27  rods  ? 


Problem  IV. 
To  find  the  area  of  a  trapezium,  or  of  an  irregular  polygon. 

13.  Divide  the  whole  figure  into  triangles,  by  draw- 
ing   DIAGONALS,  AND  FIND   THE    SUM    OF   THE  AREAS  OF  THESE 

triangles.  (Alg.  394.) 

If  the  perpendiculars  in  two  triangles  fall  upon  the  sa?ne 
diagonal,  the  area  of  the  trapezium  formed  of  the  two  trian- 
gles, is  equal  to  half  the 
product  of  the  diagonal  into 
the  sum  of  the  perpendiculars. 

Thus  the  area  of  the  trape- 
zium ABCH,  is 

iBHxALH-iBHxCM=iBHx(AL+CM.) 
Ex.  In  the  irregular  polygon  ABCDH, 


KKNSUBATION  OF  PLANE  SURFACES.  15 

(  BH=36  (  -^^=5.3 

if  the  diagonals  |  '  and  the  perpendiculars  )  CM  =9.3 

The  area=18XU.6  +  16X 7.3=379.6. 

14.  If  the  diagonals  of  a  trapezium  are  given,  the  area 
may  be  found,  nearly  in  the  same  manner  as  the  area  of  a 
parallelogram  in  Art.  5,  and  the  area  of  a  triangle  in  Art.  9. 

In  the  trapezium  ABCD,  the  sines  of  the  four  angles  at 
N,  4he  point  of  intersec- 
tion of  the  diagonals,  are 
all  equal.  For  the  two 
acute  angles  are  supple- 
ments of  the  other  two, 
and  therefore  have  the 
same  sine.  (Trig.  90.) 
Putting,  then,    sin  N  for 

the  sine  of  each  of  these  angles,  the  areas  of  the  four  tri- 
angles of  which  the  trapezium  is  composed,  are  given  by  the 
following  proportions ;  (Art.  9.) 


R  :  sin  N 


fBNxAN  :  2  arm  ABN" 
BNxCN  :  2ar«xBCN 
DNxCN  :  2  arm  CDN 

^DNxAN  :  2  arm  ADN 


And  by  addition,  (Alg.  349,  Cor.  1.)* 

R  :  8inN::BNxAN+BNxCN+DNxCN+DNxAN: 
2  area  ABCD. 

The  3d  term=(AN4-CN)x(BN+DN)=ACxBD,   by 
the  figure. 

Therefore  R  :  sin  N ::  AC  X  BD  :  2  arm  ABCD.    That  is, 
♦  Euchd,  2,  5.  Cor. 


16 


MENSURATION    OF   PLANE    SURFACES. 


As  Radius, 

To  the  sine  of  the  angle  at  the  intersection  of  the 

diagonals  of  a  trapezium  ; 
So  is  the  product  of  the  diagonals. 
To  twice  the  area  of  the  trapezium. 

It  is  evident  that  this  rule  is  applicable  to  a  parallelogram, 
as  well  as  to  a  trapezium. 

If  the  diagonals  intersect  at  right  angles,  the  sine  of  IST  is 
equal  to  radius ;  (Trig.  95.)  and  therefore  the  product  of  the 
diagonals  is  equal  to  twice  the  area.  (Alg.  356.)* 

Ex.  1.  If  the  two  diagonals  of  a  trapezium  are  3*7  and  62, 

and  if  they  intersect  at  an  angle  of  54°,  what  is  the  area  of 
the  trapezium  ?  Ans.  928. 

2.  If  the  diagonals  are  85  and  93,  and  the  angle  of  inter- 
section 74°,  what  is  the  area  of  the  trapezium  ? 


Problem  V. 

To  find  the  area  of  a  regular  polygon. 

15.  Multiply  one  of  its  sides  into  half  its  perpen- 
dicular DISTANCE  FROM  THE  CENTRE,  AND  THIS  PRODUCT 
into    the    number  of    SIDES. 

A  regular  polygon  contains  as  many  equal  triangles  as  the 
figure  has  sides.  Thus,  the  hexagon 
ABDFGH  contains  six  triangles,  each 
equal  to  ABC.  The  area  of  one  of 
them  is  equal  to  the  product  of  the 
side  AB,  into  half  the  perpendicular 
CP.  (Art.  8.)  The  area  of  the  whole, 
therefore,  is  equal  to  this  product 
multiplied  into  the  7iumler  of  sides. 


*  Euclid,  14.  5. 


XSHSURATION    OF   PLANE   SURVA0K6.  17 

E3C  1.  What  is  the  area  of  a  regular  octagon,  in  which  the 
length  of  a  side  is  60,  and  the  perpendicular  from  the  centre 
72.42G?  Ans.  17382. 

2.  What  is  the  area  of  a  regular  decagon  whose  sides  are 
46  each,  and  the  perpendicular  70.7867  ? 

16.  If  only  the  length  and  number  of  sides  of  a  regular 
polygon  be  given,  the  perpendicular  from  the  centre  may  be 
easily  found  by  trigonometry.  The  periphery  of  the  circle 
in  which  the  polygon  is  inscribed,  is  divided  into  as  many 
equal  parts  as  the  polygon  has  sides.  (Euc.  16.4.  Schol.)* 
The  arc,  of  which  one  of  the  sides  is  a  chord,  is  therefore 
known  ;  and  of  course,  the  angle  at  the  centre  subtended  by 
this  arc. 

Let  AB  be  one  side  of  a  regular  polygon  inscribed  in  the 
circle  ABDG.  The  perpendicular  CP  bisects  the  Imo  AB, 
and  the  angle  ACB.  (Euc.  3.  3.)t  Therefore,  BCP  is  the 
same  part  of  360°,  which  BP  is  of  the  perimeter  of  the 
polygon.  Then,  in  the  right  angled  triangle  BCP,  if  BP  be 
radius,  (Trig.  122.) 

R  :  BP  :  :  cot  BCP  :  CP.     That  is. 

As  Radius, 

To  half  of  one  of  the  sides  of  the  polygon ; 
So  is  the  cotangent  of  the  opposite  angle. 
To  the  perpendicular  from  the  centre. 

Ex.  1.  If  the  side  of  a  regular  hexagon  be  38  inches,  what 
is  the  area  ? 

The  angle  BCP=-iV  of  360^=30°.     Then, 

R  :  19  :  :  cot  30*»  :  32.909=CP,  the  perpendicular. 

And  the  area=10X 32.909X6=3751.6 

♦  Thomson's  Lcgendre,  2.  5.  Schol.  f  Ibid.  6.  2. 


18 


MENSURATION  OF  PLANE  SURFACES. 


2.  What  is  the  area  of  a  regular  decagon  whose  sides  are 
each  62  feet  ?  Ans.  29576. 

lY.  From  the  proportion  in  the  preceding  article,  a  table 
of  perpendiculars  and  areas  may  be  easily  formed,  for  a  series 
of  polygons,  of  which  each  side  is  a  unit.  Putting  R=l, 
(Trig.  100.)  and  w—the  number  of  sides,  the  proportion  be- 
comes 

360° 
1  t  -2  *  •  cot 1  the  perpendicular 

So  that,  the  p)erp.^=i^  cot 

2n 

And  the  area  is  equal  to  half  the  product  of  the  perpen- 
dicular into  the  number  of  sides.  (Art.  15.) 

Thus,  in  the  trigon,  or  equilateral  triangle,  the  perpendic- 
360° 


ular=^  cot. 


6 


4  cot  60°  =  0.2886752. 


And  the  area=0.4330127. 

In  the  tetragon,  or  square,  the  perpendicular=-^  cot' 


360° 

"8~ 
=-^  cot  45°  =  0.5.     And  the  area=l. 

In  this  manner,  the  following  table  is  formed,  in  which 
the  side  of  each  polygon  is  supposed  to  be  a  unit. 


A  TABLE  OF  RSGULAR  POLYGONS. 


Names. 

Sides.    Angles. 

Perpendicxilars. 

Areas. 

Trigon, 

3 

60° 

0.2886752 

0.4330127 

Tetragron, 

4 

45° 

0.5000000 

1.0000000 

Pentagon, 

5 

36° 

0.6881910 

1.7204774 

Hexagon, 

6 

30° 

0.8660254 

2.5980762 

Heptagon, 

V 

25-^ 

1.0382601 

3.6339124 

Octagon, 

8 

22i 

1.2071069 

4.8284271 

Nonag'on, 

9 

20° 

1.3737385 

6:1818242 

Decagon, 

10 

18° 

1.5388418 

7.6942088 

Undecagon, 

11 

iQ-h 

1.7028439 

9.3656399 

Dodecagon, 

12 

15° 

1.8660252 

11.1961524 

MENSURATION    OF   THE    CIRCLE.  ^.  1# 

By  this  table  may  be  calculated  the  area  of  any  other  reg- 
ular polygon,  of  the  same  number  of  sides  with  one  of  these. 
For  the  areas  of  similar  polygons  are  as  the  squares  of  their 
homologous  sides.  (Euc.  20.  0.)* 

To  find,  then,  the  area  of  a  regular  polygon,  multiply  the 
square  of  one  of  its  sides  by  tlce  area  of  a  similar  polygon  of 
which  the  side  is  a  unit. 

Ex.  1.  What  is  the  area  of  a  regular  decagon  whose  sides 
are  each  102  rods  ?  Ans.  80050.5  rods. 

2.  What  is  the  area  of  a  regular  dodecagon  whose  sides 
are  each  8Y  feet  ? 


SECTION   II. 

THE    QUADRATURE    OF    THE    CIRCLE    AND    ITS    PARTS. 

Art.  18.  Definition  I.  A  circle  is  a  plane  bounded  by  a 
line  which  is  equally  distant  in  ail  its  parts  from  a  point 
within  called  the  centre.  The  bounding  line  is  called  the 
circumference  or  periphery.  An  arc  is  any  portion  of  the 
circumference.  A  semi-circle  is  half,  and  a  quadrant  one- 
fourth  of  a  circle. 

II.  A  Diameter  of  a  circle  is  a  straight  line  drawn  through 
the  centre,  and  terminated  both  ways  by  the  circumference. 
A  Radius  is  a  straight  line  extending  from  the  centre  to  the 
circumference.  A  Chord  is  a  straidit  line  which  joins  the 
two  extremities  of  an  ar 

III.  A  Circular  Sector  is  a  spare  coniaincd  between  an 
arc  and  the  two  radii  drawn  from  the  extremities  of  the  arc. 

*  Thomson's  Legendre  1.  5.  Cor. 


20 


MENSURATION    01<'   THE    CIRCLE. 


It  may  be  less  than  a  semi-circle,  as 
AC  BO,  or  greater,  as  ACBD. 

TV.  A  Circular  Segment  is  the 
space  contained  between  an  arc  and 
its  chord,  as  ABOorABD.  The  chord 
is  sometimes  called  the  base  of  the 
segment.  The  lieight  of  a  segment 
is  the  perpendicular  from  the  middle 
of  the  base  to  the  arc,  as  PO. 

t  V.  A  Circular  Zone  is  the  space 
between  two  parallel  chords,  as 
AGHB.  It  is  called  the  middle 
zone,  when  the  two  chords  are 
equal,  as  GHDE. 


VI.  A  Circular  Ring  is  the  space  between  the  peripheries 
of  two  concentric  circles,  as  A  A',  BB'.  (Fig.  13.) 


VII.  A  Lune  or  Crescent  is  the  space  between  two  circu- 
lar arcs  which  intersect  each  other,  as  ACBD.  (Fig.  14.) 

19.  The  Squaring  of  the  Circle  is  a  problem  which  has 
exercised  the  ingenuity  of  distinguished  mathematicians  for 
many  centuries.  The  result  of  their  efforts  has  been  only 
an  approximation  to  the  value  of  the  area.  This  can  be  car- 
ried to  a  degree  of  exactness  far  beyond  what  is  necessary 
for  practical  purposes. 


MXirSURATION   OF   TBE   CIRCLE.  21 

f 

20.  If  the  circumference  of  a  circle  of  given  diameter 
were  known,  its  area  could  be  easily  found.  l\or  the  area  is 
equal  to  the  product  of  half  the  circumference  into  half  the 
diameter.  (Sup.  Euc.  5,  l.*)t  But  the  circumference  of  a 
circle  has  never  been  exactly  determined.  The  method  of 
approximating  to  it  is  by  inscribing  and  circumscribing  poly- 
gons,  or  by  some  process  of  calculation  which  is,  in  principle, 
the  same.  The  perimeters  of  the  polygons  can  be  easily 
and  exactly  determined.  That  which  is  circumscribed  is 
greater,  and  that  which  is  inscribed  is  less,  than  the  peri- 
phery of  the  circle ;  and  by  increasing  the  number  of  sides, 
the  difference  of  the  two  polygons  may  be  made  less  than 
any  given  quantity.  (Sup.  Euc.  4,  1.) 

21.  The  side  of  a  hexagon  in- 
scribed in  a  circle,  as  AB,  is  the 
chord  of  an  arc  of  60°,  and  there- 
fore equal  to  the  radius.  (Trig.  95.) 
The  chord  of  half  this  arc,  as  BO, 
is  the  side  of  a  polygon  of  12  equal 
sides.  By  repeatedly  bisecting  the 
arc,  and  finding  the  chord,  we  may 

obtain  the  side  of  a  polygon  of  an  immense  number  of  sides. 
Or  we  may  calculate  the  sine,  which  will  be  half  the  chord 
of  double  the  arc,  (Trig.  82,  cor.,)  and  the  tangent,  which 
will  be  half  the  side  of  a  similar  circumscribed  polygon. 
Thus  the  sine  AP,  is  half  of  AB,  a  side  of  the  inscribed 
hexagon ;  and  the  tangent  NO  is  half  of  NT,  a  side  of  the 
circumscribed  hexagon.  The  difference  between  the  sine 
and  the  arc  AO  is  less  than  the  difference  between  the  sine 
and  the  tangent.  In  the  section  on  the  computation  of  the 
canon,  (Trig.  223.)  by  12  successive  bisections,  begi^ping 
with  60  degrees,  an  arc  is  obtained  which  is  the  ^i\ii  of 
the  whole  circumference. 

*  In  this  manner,  the  SupplemefU  to  Play/air's  Euclid  is  referred  to  in 
this  work.  t  TtioinMn's  Legendre,  11.  5. 


22  MENSURATION    OF   THE    CIRCLE. 

The  cosine  of  this,  if  radius  be  1,  is  found  to  be  .99999996'732 
The  sine  is  .00025566546 

And  the  tangent=i^(Trig.  93.)             =.00025566347 
cosine  

The  diff.  between  the  sine  and  tangent  is  only  .00000000001 

And  the  diflference  between  the  sine  and  the  arc  is  still  less. 

Taking  then,  .000255663465  for  the  length  of  the- arc, 
multiplying  by  24576,  and  retaining  8  places  of  decimals, 
we  have  6.28318531  for  the  whole  circumference,  the  radius 
being  1.     Half  of  this, 

3.14159265 

is  the  circumference  of  a  circle  whose  radius  is  -^j  and  diam- 
eter 1. 

22.  If  this  be  multiplied  by  7,  the  product  is  21.99+or 
22  nearly.     So  that, 

Diam  :  Circum  :  :  Y  :  22,  nearly. 

If  3.14159265  be  multiplied  by  113,  the  product  is 
354.9999+,  or  355,  very  nearly.     So  that, 

Diam  ;  Circum  ;  :  113  :  355,  very  nearly. 

The  first  of  these  ratios  was  demonstrated  by  Archimedes. 

There  are  various  methods,  principally  by  infinite  series 
and  fluxions,  by  which  the  labor  of  carrying  on  the  approx- 
imation to  the  periphery  of  a  circle  may  be  very  much 
abridged.  The  calculation  has  been  extended  to  nearly  150 
places  of  decimals.  But  four  or  five  places  are  sufl&cient 
for  most  practical  purposes. 

After  determining  the  ratio  between  the  diameter  and  the 
circumference  of  a  circle,  the  following  problems  are  easily 
solved. 


MENSURATION    OF   THE    CIRCUS.  2S 

Problem  I. 
To  find  the  circitmference  of  a  circle  from  its  diameter. 

23.  Multiply  THE  diameter  by  3*1 41 59*^ 

Or, 

Multiply  the  diameter  hy  22  and  divide  the  product  hy  *l.  Or, 
multiply  the  diameter  by  355,  and  divide  the  product  by 
113.  (Art.  22.) 

Ex.  1.  If  the  diameter  of  the  earth  be  ^930  miles,  what 
is  the  circimiference  ?  Ans.  249128  miles. 

2.  How  many  miles  does  the  earth  move,  in  revolving 
round  the  sun ;  supposing  the  orbii  to  be  a  circle  whose 
diameter  is  190  million  miles  ?  Ans.  596,902,100. 

8.  What  is  the  circumference  of  a  circle  whose  diameter 
is  V69843  rods  ? 

Problem  II. 
To  find  the  diameter  of  a  circle  from  its  circumference, 

24.  Divide  the  circumference  by  3.1 41 59* 

Or, 

Multiply  the  circumference  by  Y,  and  divide  the  product  by 
22.  Or,  multiply  the  circumference  by  113,  and  divide 
the  product  by  355.  (Art.  22.) 

Ex.  1.  If  the  circumference  of  the  sun  be  2,800,000  miles, 
what  is  his  diameter?  Ans.  891,267. 

2.  What  is  the  diameter  of  a  tree  which  is  5^  feet  round  ? 

25.  As  multiplication  is  more  easily  performed  than  divis- 
ion, there  will  be  an  advantage  in  exchanging  the  divisor 


*  In  many  caset,  3.1416  will  be  sufficiently  accurate. 


24  MENSURATION    OF    THE    CIRCLE. 

3.14159  for  a  multiplier  which  ■will  give  the  same  result. 
In  the  proportion 

3.14159  :  1  :  :  Circum  :  Diam. 

to  find  the  fourth  term,  we  may  divide  the  second  by  the 
first,  and  multiply  the  quotient  into  the  third.  Xow,  1-^ 
3.14159=0.31831.  If,  then,  the  circumference  of  a  circle 
be  multiplied  by  .31831,  the  product  will  be  the  diameter. 

Ex.  1.  If  the  circumference  of  the  moon  be  6850  miles, 
what  is  her  diameter  ?  Ans.   2180. 

2.  If  the  whole  extent  of  the  orbit  of  Saturn  be  5650 
million  miles,  how  far  is  he  from  the  siin  ? 

3.  If  the  periphery  of  a  wheel  be  4  feet  7  inches,  what  is 
its  diameter? 

Problem  III. 
To  find  the  length  of  an  arc  of  a  circle. 

26.  As  360°,  to  the  number  of  degrees  in  the  arc  ; 

So  is  the  circumference  of  tlie  circle,  to  the  length  of  the  arc. 

The  circumference  of  a  circle  being  divided  into  360°jflH|L 
(Trig.  73.)  it  is  evident  that  the  length  of  an  arc  of  any  less 
number  of  degrees  must  be  a  proportional  part  of  the  whole. 

Ex.  What  is  the  length  of  an  arc  of  16°,  in  a  circle  whose 
radius  is  50  feet  ? 

The  circumference  of  the  circle  is  314.159  feet.  (Art.  23.) 

Then  360  :  16  :  :  314.159  :  13.96  feet. 

2.  If  we  are  95  millions  of  miles  from  the  sun,  and  if  the 
earth  revolves  round  it  in  365i  days,  how  far  are  we  carried 
in  24  hours?  Ans.  1  million  634  thousand  miles. 

27.  The  length  of  an  arc  may  also  be  found,  by  multiply- 
ing the  diameter  into  the  number  of  degrees  in  the  arc,  and 


MBNSURATION    OF   THE   CIRCIJB. 


25 


this  product  into  .0087266,  which  is  the  length  of  one  de- 
gree, in  a  circle  whose  diameter  is  1.  For  3,14169-r360= 
0.00872  60.  And  in  different  circles,  the  circumferences,  and 
of  course  the  degrees,  are  as  the  diameters.  (Sup.  Euc.  8,  1.)* 

Ex.  1.  What  is  the  length  of  an  arc  of  10°  16'  in  a  circle 
whose  radius  is  68  rods  ?  Ans.  12.165  rods. 

2.  If  the  circumference  of  the  earth  be  24913  miles,  what 
is  the  length  of  a  degree  at  the  equator  ? 

28.  The  length  of  an  arc  is  frequently  required,  when 
the  number  of  degrees  is  not  given.  But  if  the  radius  of  the 
circle,  and  either  the  cliord  or  the  fieight  of  the  arc,  be 
known  ;  the  number  of  degrees 
may  be  easily  found. 

Let  AB  be  the  chord,  and  PO 
the  height,  of  the  arc  AOB.  As 
the  angles  at  P  are  right  angles,  and 
AP  is  equal  to  BP;  (Art.  18.  Def. 
4.)  AO  is  equal  to  BO.  (Euc.  4, 
l.)t    Then, 

BP  is  the  sine,  and  CP  the  cosine,       )  ^^  ^^ .  ^j^^  ^^  ^^^ 
OP  the  versed  sine,  and  BO  the  chord,  ) 


And  in  the  right  angled  triangle  GBP, 

PR  ♦  R       i  ^1*  •  sin  BCP  or  BO. 
*  i  CP  :  cosBCPorBO. 

Ex.  1.  If  the  radius  CO=25,  and  the  chord  AB=43.3; 
what  is  the  length  of  the  arc  AOB  ? 

CB  :  R  :  :  BP  :  sin  BCP  or  BO— 60'' very  nearly. 

The  circumference  of  the  circle  -"3.14159  X50-»167.08. 

Ajid  360° :  60° : :  157.08 :  26.18=OB.  Therefore,AOB=62.36. 


*  Thomson's  Legendre,  10.  b. 


t  Ibid.,  5.  1. 


26 


MENSURATION    OF    THE    CIRCLE. 


/        2.  What  is  the  length  of  an  arc  whose  chord  is  216^,  in  a 
circle  whose  radius  is  126?  Ans.  261.8. 

29.  If  only  the  chord  and  the  height  of  an  arc  be  given, 
the  radius  of   the  circle   may   be 
found,  and  then  the  length  of  the 
arc. 

If  BA  be  the  chord,  and  PO  the 
the  height  of  the  arc  AOB,  then 
(Euc.  35.  3.)* 


DP=Jl_.      And  DO=OP+DP=OP+S--' 
OP  OP 

That  is,  the  diameter  is  equal  to  the  height  of  the  arc,  + 
the  square  of  half  the  chord  divided  by  the  height. 

The  diameter  being  found,  the  length  of  the  arc  may  be 
calculated  by  the  two  preceding  articles. 

Ex.  1.  If  the  chord  of  an  arc  be  173.2,  and  the  heischt  50, 
what  is  the  length  of  the  arc  ? 


The   diameter  =50+5^=200.    The  arc  contains  120^ 
50 

(Art.  28.)  and  its  length  is  209.44.  (Art.  26.) 

2.  What  is  the  length  of  an  arc  whose  chord  is  120,  and 
height  45?  Ans.   160.8. 


Problem  IV. 

To  find  the  area  of  a  circle. 

30.  Multiply    the    square    of   the    diameter    by  the 
decimals  .7854. 


*  Thomson's  Legendre,  10,  5. 


lOBMntATION    OF   THE    CIRCLE.  %% 

Or, 

Multiply  half  the  diameter  into  half  the  circum- 
ference. Or,  multiply  the  whole  diameter  into  the  whole 
circumference,  and  take  -^  of  the  product. 

The  area  of  a  circle  is  equal  to  the  product  of  half  the 
diameter  into  half  the  circumference ;  (Sup.  Euc,  5,  1.)  or, 
which  is  the  same  thing,  ^  the  product  of  the  diameter  and 
circumference.  If  the  diameter  be  1,  the  circumference  is 
3.14159;  (Art.  23.)  one-fourth  of  which  is  0.7854  nearly. 
But  the  areas  of  different  circles  are  to  each  other,  as  the 
squares  of  their  diameters.  (Sup.  Euc.  Y,  1.)*  The  area  of 
any  circle,  therefore,  is  equal  to  the  product  of  the  square  of 
its  diameter  into  0.7854,  which  is  the  area  of  a  circle  whose 
diameter  is  1. 

Ex.  1.  What  is  the  area  of  a  circle  whose  diameter  is  623 
feet^.  Ans.  304836  square  feet. 

2.  How  many  acres  are  there  in  a  circular  island  whose 
diameter  is  124  rods.  Ans.   75  acres,  and  76  rods. 

3.  If  the  diameter  of  a  circle  be  113,  and  the  circumfer- 
ence 355,  what  is  the  area  ?  Ans.  10029. 

4.  How  many  square  yards  are  there  in  a  circle  whose 
diaraelcr  is  7  feet  ? 

81.  If  the  circumference  of  a  circle  be  given,  the  area  may 
be  obtained,  by  fir-  the  diameter ;  or,  without  finding 

the  diameter,  by  mu  ^  y  -  the  square  of  the  circumference 
by  .07958. 

For,  if  the  circumference  of  a  circle  be  1,  tlie  diameter  = 
1-^3.14159=0.31831  ;  and  |  the  product  of  this  into  the 
circumference  is  .07958  the  area.  But  the  areas  of  different 
circles,  being  as  the  squares  of  their  diameters,  are  also  as 
the  squares  of  their  circumferences.  (Sup.  Euc.  8,  1.) 


•  Thomson's  Legendre,  28.  4.   Cor. 


28 


MENSURATION    OF   THE    CIRCLE. 


Ex.  1.  If  the  circumference  of  a  circle  be  136  feet,  what 
is  the  area  ?  Ans.   1472  feet. 

2.  What  is  the  surface  of  a  circular  fish-pond,  which  is 
10  rods  in  circumference  ? 

32.  If  the  area  of  a  circle  be  given,  the  diameter  may  be 
found,  by  dividing  the  area  by  .7854,  and  extracting  the 
square  root  of  the  quotient. 

This  is  reversing  the  rule  in  Art.  30. 

Ex.  1.  What  is  the  diameter  of  a  circle  whose  area  is 
380.1336  feet? 

Ans.  380.1336H-.'7854  =  484.     And  V4'84  =  22. 

2.  What  is  the  diameter  of  a  circle  whose  area  is  19.635  ? 

33.  The  area  of  a  circle,  is  to  the  area  of  the  circumscribed 
square ;  as  .7854  to  1  ;  and  to  that  of  the  inscribed  square 
as  .7854  to  i. 

Let  ABDF  be  the  inscribed 
square,  and  LMNO  the  circum- 
scribed square,  of  the  circle  ABDF. 
The  area  of  the  circle  is  equal  to 
AD"' X. 7854.  (Art.  30.)  But  the 
area  of  the  circumscribed  square 

(Art.  4.)  is  equal  to  ON^=AD^. 
And  the  smaller  square  is  half  of 
the  larger  one.  For  the  latter  con- 
tains 8  equal  triangles,  of  which  the  former  contains  only  4. 

Ex.  What  is  the  area  of  a  square  inscribed  in  a  circle 
whose  area  is  159  ?         Ans.  .7854  :  i  :  :  159  :  101.22. 


c        \^ 

12 


Problem  V. 
To  find  the  area  of  a  sector  of  a  circle. 
34.    Multiply  the    radius  into    half   the  length    of 

THE    ARC. 


MENSURATION    OF   THE    CIRCLE. 


Or, 

As  360,  TO  THE  WUMBKR    OF    DEGREES    IN   THE    ARC  ,* 
So    IS    THE    AREA    OF   THE    CIRCLE,    TO     THE     AREA     OF   THE 
SECTOR. 

It  is  evident,  that  the  area  of  the  sector  has  the  same 
ratio  to  the  area  of  the  circle,  which  the  length  of  the  arc 
has  to  the  length  of  the  whole  circumference ;  or  which  the 
number  of  degrees  in  the  arc  has  to  the  number  of  degrees 
in  the  circumference. 

Ex.  1.  If  the  arc  AOB  be  120°, 
and  the  diameter  of  the  circle  226  ; 
what  is  the  area  of  the  sector 
AOBC? 


The  area  of  the  whole  circle   is 
40116.  (Art.  30.) 

And   360°   :    120°  :  :  40115    : 
133 71 1,  the  area  of  the  sector. 

2.  AVhat  is  the  area  of  a  quadrant  whose  radius  is  621  ? 

3.  What  is  the  area  of  a  semi-circle,  whose  diameter  is 
328? 

4.  What  is  the  area  of  a  sector  which  is  less  than  a  semi- 
circle, if  the  radius  be  15,  and  the  chord  of  its  arc  12  ? 

Half  the  chord  is  the  sine  of  23°  34|'  nearly.  (Art.  28.) 

The  whole  arc,  then,  is  47°    9^' 

The  area  of  the  circle  is  706.86 

And  360°  :  47°  9i' : :  706.86  :  92.6  the  area  of  the  sector. 

5.  If  the  arc'ADB  be  240  degrees,  and  the  radius  of  the 
circle  113,  what  is  the  area  of  the  sector  ADBC  ? 


Problem  VI. 
To  finl  the  area  of  a  segment  of  a  circle. 
2o.    Find   the    aeba  of   the    sector  which    has    the 


80  MENSURATION"    OF   THE    CIRCLE. 

SAME  ARC,  AND  ALSO  THE  AREA  OF  A  TRIANGLE  FORMED  BY 
THE  CHORD  OF  THE  SEGMENT  AND  THE  RADII  OF  THE  SEC- 
TOR. 

Then,  if  the  segment  be  less  than  a  semi-circle, 
subtract  the  area  of  the  triangle  from  the  area  of 
the  sector.      but,  if  the   segment  be    greater  than  a 

SEMI-CIRCLE,  ADD  THE  AREA  OF  THE  TRIANGLE  TO  THE  AREA 
OF    THE    SECTOR. 

If  the  triangle  ABC,  be  taken 
from  the  sector  AOBC,  it  is  evi- 
dent the  difference  will  be  the  seg- 
ment AOBP,  less  than  a  semi-cir- 
cle. And  if  the  same  triangle  be 
added  to  the  sector  ADBC,  the 
sum  will  be  the  segment  ADBP, 
greater  than  a  semi-circle. 

The  area  of  the  triangle  (Art.  8.) 
is  equal  to  the  product  of  half  the  chord  AB  into  CP,  which 
is  the  difference  between  the  radius  and  PO  the  height  of  the 
segment.  Or  CP  is  the  cosine  of  half  the  arc  BOA.  If  this 
cosine  and  the  chord  of  the  segment  are  not  given,  they 
may  be  foimd  from  the  arc  and  the  radius. 

Ex.  1.  If  the  arc  AOB  be  120°,  and  the  radius  of  the 
circle  be  113  feet,  what  is  the  area  of  the  segment  AOBP  ? 

In  the  right  angled  triangle  BCP, 

R  :  BC  :  :  sin  BCO  :  BP=97.86,  half  the  chord.  (Art.  28.) 

The  cosine  PC=i  CO  (Trig.  96,  Cor.)  =56.5 

The  area  of  the  sector  AOBC  (Art.  34.)  =13371.67 

The  area  of  the  triangle  ABC=BPxPC  =   5528.97 

The  area  of  the  segment,  therefore,  =   7842.7 

2.  If  the  base  of  a  segment,  less  than  a  semi-circle,  be  10 


MENSURATION    OF    THE    CIRCLE.  $X 

feet,  and  the  radius  of  tlie  circle   12  feet,  what  is  the  area 
of  the  segment  ? 

The  arc  of  the  segment  contains     49-J-     degrees.  (Art.  28.) 
The  area  of  the  sector  =61.89  (Art.  34.) 

The  area  of  the  triangle  =54.54 

And  the  area  of  the  segment      =  7.35  square  feet. 

3.  What  is  the  area  of  a  circular  segment,  whose  height 
is  19.2  and  base  70  ?  Ans.  947.86. 

4.  What  is  the  area  of  the  segment  ADBP,  (Fig.  9.)  if 
the  base  AB  be  195.7,  and  the  height  PD  169.5  ? 

Ans.  32272.    j 

36.  The  area  of  any  figure  which  is  bounded  partly  by 
arcs  of  circles,  and  partly  by  right  lines,  may  be  calculated, 
by  finding  the  areas  of  the  segments  under  the  arcs,  and  then 
the  area  of  the  rectilinear  space  between  the  chords  of  the 
arcs  and  the  other  right  lines. 

Thus,  the  Gothic  arch  ACB, 
contains  the  two  segments 
AC II,  BCD,  and  the  plane  tri- 
angle ABC. 

Ex.  If  AB  be  110,  each  of 
the  fines  AC  and  BC  100,  and 
the  height  of  each  of  the  seg- 
ments ACH,  BCD  10.435; 
what  is  the  area  of  the  whole  figure  ? 

The  areas  of  the  two  segments  are  1404 

The  area  of  the  triangle  ABC  is  4593.4 

And  the  whole  figure  is  5997.4 


32 


MENSURATION    OF   THE    CIRCLE. 


Problem  VII. 
To  find  the  area  of  a  circular  zone, 
37.  From    the  area    of    the  whole    circle,  subtract 

THE    TWO    segments    ON   THE    SIDES    OF   THE    ZONE. 

If  from  the  whole  circle  there  be  taken  the  two  segments 
ABC  and  DFH,  there  will  remain 
the  zone  ACDH. 

Or,  the  area  of  the  zone  may  be 
found  by  subtracting  the  segment 
ABC  from  the  segment  HBD  :  Or, 
by  adding  the  two  small  segments 
GAH  and  VDC,  to  the  trapezoid 
ACDH.  (See  Art.  36.) 

The  latter  method  is  rather  the 
most  expeditious  in  practice,  as  the  two  segments  at  the  end 
of  the  zone  are  equal. 

Ex.  1.  What  is  the  area  of  the  zone  ACDH,  if  AC  is 
7.75,  DH  6.93,  and  the  diameter  of  the  circle  8  ? 


The  area  of  the  whole  circle  is 
of  the  segment  ABC 
of  the  segment  DFH 
of  the  zone  ACDH 


17.32 

9.82 


50.26 


23.12 


2.  What  is  the  area  of  a  zone,  one  side  of  which  is  23.25, 
and  the  other  side  20.8,  in  a  circle  whose  diameter  is  24  ? 

Ans.  208. 

38.  If  the  diameter  of  the  circle  is  not  given,  it  may  be 
found  from  the  sides  and  the  breadth  of  the  zone. 

Let  the  centre  of  the  circle  be  at  0.  Draw  ON  perpen- 
dicular to  AH,  NM  perpendicular  to  LR,  and  HP  perpen- 
dicular to  AL.     Then, 


MENSURATION    OF    THE    CIRCLE.  33 

AN=iAH,  (Euc.  3.  3.)*        MN=KLA+RH) 
LM=iLR,  (Euc.  2.  6.)t        PA=LA— RH. 

The  triangles  API!  and  OMN  are  siniilar,  because  the 
sides  of  one  are  perpendicular  to  those  of  the  other,  each  to 
each.     Therefore, 

PII  :  PA  : :  MN  :  MO 
MO  being  found,  we  have  ML — MO=OL. 


And  the  radius  CO=vOL»+CL*.  (Euc.  41.  l.)l 

Ex.  If  the  breadth  of  the  zone  ACDH  (Fig.  12.)  be  6.4, 
and  the  sides  6.8  and  6 ;  what  is  the  radius  of  the  circle  ? 

PA=3.4— 3=0.4.         And,  MN=i(3.4+3)=3.2. 
Then,  6.4  :  0.4  : :  3.2  :  0.2=M,O.     And,  3.2— 0.2=3=OL 
And  the  radius  CO=V3'4-(3.4)'=4.534. 

^^  Problem  VIII. 

To  find  the  area  of  a  lune  or  crescent. 
39.  Find  the  difference  of  the  two  segments  which 

ARE    between    the    ARCS    OF   THE    CRESCENT    AND    ITS    CHORD. 

If  the  segment  ABC,  be 
taken  from  the  segment  ABD  ; 
there  will  remain  the  lune  or 
crescent  ACBD. 

Ex.  If  the  chord  AB  be  8B, 
the  height  CH  20,  and  the 
height  DH  40 ;  what  is  the 
area  of  the  crescent  ACBD  ? 

The  area  of  the  segment  ABD  is  2698 

of  the  segment  ABC  1220 

of  the  crescent  ACBD  1478 


•  Thomson's  Lcgendre,  6.  2.        f  ^^  1^-  4.        t  ^^'^  ^l-  4* 
2» 


84  MENSURATION    OF   THE    CIRCLE. 


Problem  IX. 

To  find  the  area  of  a  ring,  included  between  the  periphe- 
ries of  two  concentric  circles. 

40.    Find  tsk    difference    op  the  areas  of  the  two 

CIRCLES. 

Or, 

Multiply  the  product  of  the  sum  and  diflference  of  the  two 
diameters  by  .TSS^. 

The  area  of  the  ring  is  evidently  equal  to  the  diflference 
between  the   areas   of  the   two  cir- 
cles AB  and  A'B'. 

But  the  area  of  each  circle  is  equal 
to  the  square  of  its  diameter  multi- 
plied into  .78.54.  (Art.  30.)  And 
the  difference  of  these  squares  is  equal 
to  the  product  of  the  sum  and  dif- 
ference of  the  diameters.  (Alg.  191.)  Therefore  the  area  of 
the  ring  is  equal  to  the  product  of  the  sum  and  diflference 
of  the  two  diameters  multiplied  by  .'7854. 

Ex.  1.  If  AB  be  221,  and  A'B'  106,  what  is  the  area  of 
the  ring?      Ans.  (22p X. 7854)— (106=  X.'7854)  =  29535. 

2.  If  the  diameters  of  Saturn's  larger  ring  be  205,000 
and  190,000  miles,  how  many  square  miles  are  there  on  one 
side  of  the  ring  ?  ^ 

Ans.  395000X15000X. 7854=4,653,495,000. 

PROMISCUOUS    EXAMPLES    OF    AREAS. 

Ex.  1.  What  is  the  expense  of  paving  a  street  20  rods 
long  and  2  rods  wide,  at  5  cents  for  a  square  foot  ? 

Ans.  544-^  dollars. 


MKN8URATI0N   OF   THE    CIRCLE.  35 

2.  If  an  equilateral  triangle  contains  as  many  square  feet 
as  there  are  inches  in  one  of  its  sides ;  what  is  the  area  of 
the  trijingle  ? 

Let  ;2;=the  number  of  square  feet  in  the  area. 

X 

Then— •= the  number  of  linear  feet  in  one  of  the  sides. 

And,  (Art.  11.)  ar=i/-:^yx  V3=_^XV3. 
\12/  676 

Reducing  the  equation,  a;=_  =332.55  the  area. 
V3 

3.  What  is  the  side  of  a  square  whose  area  is  equal  to  that 
of  a  circle  452  feet  in  diameter? 


Ans.  V(452)»X. 7854  =  400.574.  (Arts.  30  and  7.) 

4.  What  is  the  diameter  of  a  circle  which  is  equal  to  a 
square  whflie  side  is  36  feet  ? 

Ans.  V('307ToT7854^4O.62l7.  (Arts.  4  and  32.) 

5.  What  is  the  area  of  a  square  inscribed  in  a  circle  whose 
diameter  is  132  feet  ? 

/gf^^'  8712  square  feet.  (Art.  33.) 

6.  How  much  carpeting,  a  yard  wide,  will  be  necessary  to 
cover  the  floor  of  a  room  which  is  a  regular  octagon,  the 
sides  being  eight  feet  each  ?  Ans.  34f  yards. 

7.  If  the  diagonal  of  a  square  be  16  feet,  what  is  the 
area?  Ans.  128  feet.  (Art.  14.) 

8.  If  a  carriage-wheel  four  feet  in  diameter  revolve  300 
times,  in  going  round  a  circular  green ;  what  is  the  area  of 
the  green  ? 

Ans.  4164i  sq.  rods,  or  26  acres,  3  qrs.  and  34^  rods. 

9.  What  will  be  the  expense  of  papering  the  sides  of  a 
room,  at  10  cents  a  square  yard ;  if  the  room  be  21  feet  long, 


3G 


*  MENSURATION    OF   THE    CIRCLE. 

4 


18  feet  broad,  and  12  feet  high ;  and  if  there  be  deducted  3 
windows,  each  5  feet  by  3,  two  doors  8  feet  by  4^,  and  one 
fire-place  6  feet  by  4i  ?  Ans.  8  dollars  80  cents. 

10.  If  a  circular  pond  of  water  10  rods  in  diameter  be 
surrounded  by  a  gravelled  walk  8}  feet  wide ;  what  is  the 
area  of  the  walk?  Ans.  16^  sq.  rods.  (Art.  40.) 


11.  If  CD,  the  base  of  the 
isosceles  triangle  VCD,  be  60 
feet,  and  the  area  1200  feet; 
and  if  there  be  cut  off,  by  the 
line  LG  parallel  to  CD,  the  tri- 
angle VLG,  whose  area  is  432 
feet ;  what  are  the  sides  of  the 
latter  triangle  ? 

Ans.  30,  30,  and  36  feet. 


12.  What  is  the  area  of  an  equilateral  triangle  inscribed 
in  a  circle  whose  diameter  is  52  feet  ? 

Ans.  878.15  sq.  ft. 

13.  If  a  circular  piece  of  land  is  inclosed  by  a  fence,  in 
which  10  rails  make  a  rod  in  length;  and  if  the  field  con- 
tains as  many  square  rods,  as  there  are  rails  in  the  fence ; 
what  is  the  value  of  the  land  at  120  dollars  an  acre  ? 

Ans.  942.48  dollars. 

14.  If  the  area  of  the  equilat- 
eral triangle  ABD  be  219.5375 
feet ;  what  is  the  area  of  the  cir- 
cle OBDA,  in  which  the  triangle 
is  inscribed? 
The  sides  of  the  triangle  are  each 

22.5167.   (Art.  11.) 

And  the  area  of  the  circle  is  630.93. 


MENSURATION    OF    SOLIDS.    *  87 

15.  If  6  concentric  circles  are  so  drawn,  that  the  space 
between  the  least  or  1st,  and  the  2d   is  21.2058, 

between  the  2d    and  the  3d   is  35.343, 

between  the  3d    and  the  4th  is  49.4802,' 

between  the  4th  and  the  5th  is  63.6174, 

between  the  5th  and  the  6th  is  V7.Y546; 

what  Ihre  the  several  diameters,  supposing  the  longest  to  be 

equal  to  6  times  the  shortest  ? 

Aus.  3,  6,  0,  12,  15,  and  18. 

16.  If  the  area  between  two  concentric  circles  be  1202.64 
square  inches,  and  the  diameter  of  the  lesser  circle  be  19 
uiche8»t>what  is  the  diameter  of  the  other  ? 

17.  What  is  the  area  of  a  circular  segment,  whose  height 
is  9,  and  base  24? 


SECTION   III. 

SOLIDS    BOUNDED    BY    PLANE    SURFACES. 

Abt.  41.  DEFINr^(0||^  I.  A  prism  is  a  solid  bounded  by 
plane  figures  or  faces,  two  of  which  are  parallel,  similar,  and 
equal ;  and  the  others  are  parallelograms. 

II.  The  parallel  planes  are  sometimes  called  the  bases  or 
ends;  and  the  other  figures  the  sides  of  the  prism.  The 
latter  taken  together  constitute  the  lateral  surface, 

III.  A  prism  is  riffht  or  oblique,  according  as  the  sides  are 
perpendicular  or  oblique  to  the  bases. 

IV.  The  height  of  a  prism  is  the  perpendicular  distance 
between  the  planes  of  the  bases.  In  a  right  prism,  there- 
fore, tLe  height  is  equal  to  the  length  of  one  of  the  sides. 

V.  A  Parallelopiped  is  a  |)ri8ni  whose  bases  are  parallelo- 
grams. 


38 


MENSURATION    OF   SOLIDS. 


VI.  A  Cuhe  is  a  solid  bounded  by  six  equal  squares.  It 
is  a  right  prism  whose  sides  and  bases  are  all  equal. 

VII.  A  Pyramid  is  a  solid  bounded  by  a  plane  figure 
called  the  base,  and  several  triangular  planes,  proceeding  from 
the  sides  of  the  base,  and  all  terminating  in  a  single  point. 
These  triangles  taken  together  constitute  the  lateral  surface. 

VIII.  A  pyramid  is  reguUar,  if  its  base  is  a  regular  poly- 
gon, and  if  a  line  from  the  centre  of  the  base  to  the  vertex 
of  the  pyramid  is  perpendicular  to  the  base.  This  line  is 
called  the  axis  of  the  pyramid. 

IX.  The  height  of  a  pyramid  is  the  perpendicular  distance 
from  the  summit  to  the  plane  of  the  base.  In  a  regular  pyr- 
amid, it  is  the  length  of  the  axis. 

X.  The  slant-height  of  a  regular  pyramid,  is  the  distance 
from  the  summit  to  the  middle  of  one  of  the  sides  of  the  base. 

XI.  A  frustum  or  trunk  of  a  pyramid  is  a  portion  of  the 
solid  next  the  base,  cut  off  by  a  plane  parallel  to  the  base. 
The  height  of  the  frustum  is  the  perpendicular  distance 
of  the  two  parallel  planes.  The  slant-height  of  a  frus- 
tum of  a  regular  pyramid,  is  the  distance  from  the  middle  of 
one  of  the  sides  of  the  base,  to  the  middle  of  the  corres- 
ponding side  in  the  plane  above.  It  is  a  line  passing  on 
the  surface  of  the  frustum,  through  the  middle  of  one  of 
its  sides. 

XII.  A  Wedge  is  a  soUd  of  five  sides,  viz.  a  rectangular 
base,  two  rhomboidal 

sides  meeting  in  an 
edge,  and  two  tri- 
angular ends ;  as 
ABHG.  The  base 
is  ABCD,  the  sides 
are  ABHG  and 
DCHG,  meeting  in 
the  edge  GH,  and 
the  ends  are  BOH  and  ADG.     The  height  of  the  wedge  is  a 


MSNSURATION    OF   SOLIDS.  39 

perpendicular  drawn  from  any  point  in  the  edge,  to  the  plane 
of  the  base,  as  GP. 

XIII.  A  Prismoid  is  a  solid  whose  ends  or  bases  are  par- 
allel, but  not  similar,  and  whose  sides  are  quadrilateral.  It 
differs  from  a  prism  or  a  frustum  of  a  pyramid,  in  having  its 
ends  dissimilar.  It  is  a  rectangular  prismoid,  when  its  ends 
are  right  parallelograms. 

XIV.  A  linear  side  or  edge  of  a  solid  is  the  line  of  intersec- 
tion of  two  of  the  planes  which  form  the  surface. 

42.  The  common  measuring  unit  of  solids  is  a  cube,  whose 
sides  are  squares  of  the  same  name.  The  sides  of  a  cubic 
inch  are  square  inches ;  of  a  cubic  foot,  square  feet,  &c. 
Finding  the  capacity,  solidity*  or  solid  contents  of  a  body,  is 
fmding  the  number  of  cubic  measures,  of  some  given  de- 
nomination contained  in  the  body. 

In  solid  mectsure. 

1728     cubic  incha  =1  cubic  foot, 
2*7     cubic  feet      =1  cubic  yard, 
4492i  cubic  feet      =1  cubic  rod, 
32768000     cubic  rods      =1  cubic  mile, 
282     cubic  inches  =1  ale  gallon, 
231     cubic  inches  =1  wine  gallon, 
2150.42     cubic  inches  =1  bushel, 

1     cubic  foot  of   pure  water  weighs  1000 
avoirdupois  ounces,  or  62i  pounds. 

Problkm  I. 
To  find  the  soLmnr  of  a  prism. 

48.    MlTLTIPLY  THE  AREA  OF  THE  BASE  BY  THE  HEIGHT. 

This  is  a  general  rule,  applicable  to  parallelopipeds 
whether  right  or  obhque,  cubes,  triangular  prisms,  <fec. 

♦  Sec  note  A. 


40  MENSURATION    OF   SOLIDS. 

As  surfaces  are  measured,  by  comparing  them  with  a  right 
parallelogram  (Art.  3.) ;  so  solids  are  measured,  by  com- 
paring them  with  a  right  parallelepiped. 

If  ABCD  be  the  base  of  a  right  ^ 
parallelopiped,  as  a  stick  of  timber 
standing  erect,  it  is  evident  that  the 
number  of  cubic  feet  contained  in  one 
foot  of  the  height,  is  equal  to  the 
mimber  of  square  feet  in  the  area  of 
the  base.     And  if  the  sohd  be  of  any 

other  height,  instead  of  one  foot,  the  contents  must  have  the 
same  ratio.  For  parallelepipeds  of  the  same  base  are  to 
each  other  as  their  heights.  (Sup.  Euc.  9.  3.)*  The  solidity 
of  a  right  parallelopiped,  therefore,  is  equal  to  the  product 
of  its  length,  breadth,  and  thickness.     See  Alg.  397. 

And  an  oblique  parallelopiped  being  equal  to  a  right  one 
of  the  same  base  and  altitude,  (Sup.  Euc.  7.  3)f  is  equal  to 
the  area  of  the  base  multiplied  into  the  perpendicular  height. 
This  is  true  also  of  prisms,  whatever  be  the  form  of  their 
bases.  (Sup.  Euc.  2.  Cor.  to  8, 3.  Thomson's  Legendre,  12.  7.) 

44.  As  the  sides  of  a  cube  are  all  equal,  the  solidity  is 
found  by  cubing  one  of  its  edges.  On  the  other  hand,  if  the 
solid  contents  be  given,  the  length  of  the  edges  may  be 
found,  by  extracting  the  cube  root. 

45.  When  solid  measure  is  cast  by  Duodecimals,  it  is  to 
be  observed  that  inches  are  not  primes  of  feet,  but  thirds. 
If  the  unit  is  a  cubic  foot,  a  solid  which  is  an  inch  thick  and 
a  foot  square  is  a  prime  ;  a  parallelopiped  a  foot  long,  an 
inch  broad,  and  an  inch  thick  is  a  second,  or  the  twelfth  part 
of  a  prime  ;  and  a  cubic  inch  is  a  third,  or  the  twelfth  part 
of  a  second.  A  linear  inch  is  -i\  of  a  foot,  a  square  inch  -ri t 
of  a  foot,  and  a  cubic  inch  tt^-s  of  a  foot. 

*  Thomson's  Legendre,  9.  7.  f  Ibid.,  7.  7. 


MKNSURATION    OP   SOLIDS.  41 

Ex.  1.  What  are  the  solid  contents  of  a  stick  of  tlmbe^ 
which  is  31  feet  long,  1   foot  3  inches  broad,  and  9  inches 
thick?  Ans.  29  feet  9",  or  29  feet  108  inches. 

2.  What  is  the  solidity  of  a  wall  which  is  22  feet  long,  12 
feet  high,  and  2  feet  6  inches  thick  ? 

Ans.  660  cubic  feet. 

3.  What  is  the  capacity  of  a  cubical  vessel  which  is  2  feet 
3  inches  deep  ? 

Ans.  11  F.  4'  8"  3'",  or  11  feet  6*75  inches. 

4.  If  the  base  of  a  prism  be  108  square  inches,  and  the 
height  30  feet,  what  are  the  solid  contents  ? 

Ans.  27  cubic  feet. 

6.  If  the  height  of  a  square  prism  be  2\  feet,  and  each 
side  of  the  base  10^  feet,  what  is  the  solidity  ? 

The  area  of  the  base  =   10ixl0i==106f  sq.  feet. 
And  the  solid  contents  =  106iX   2-i-=240i  cubic  feet. 

6.  If  the  height  of  a  prism  be  23  feet,  and  its  base  a  reg- 
ular pentagon,  whose  perimeter  is  18  feet,  what  is  the  so- 
lidity ?  Ans.  512.84  cubic  feet. 

46.  The  number  of  gallons  or  bushels  which  a  vessel  will 
contain  may  be  found,  by  calculating  the  capacity  in  inches, 
and  then  dividing  by  the  number  of  inches  in  1  gallon  or 
bushel. 

The  weight  of  water  in  a  vessel  of  given  dimensions  is 
easily  calculated ;  as  it  is  found  by  experiment,  that  a  cubic 
foot  of  pure  water  weighs  1000  ounces  avoirdupois.  For 
the  weight  in  ounces,  then,  multiply  the  cubic  feet  by  1000 ; 
or  for  the  weight  in  pounds,  multiply  by  62^. 

Ex.  1.  How  many  ale  gallons  are  there  in  a  cistern  whicli 
is  1 1  feet  9  inches  deep,  and  whose  base  is  4  feet  2  inches 
square? 


42  MENSURATION    OF   SOLIDS. 

•"  The  cistern  contains  352500  cubic  inches; 

And  352500-7-282  =  1250. 

2.  How  many  wine  gallons  will  fill  a  ditch  3  feet  11  inches 
wide,  3  feet  deep,  and  462  feet  long  ?  Ans.  40608. 

3.  What  weight  of  water  can  be  put  into  a  cubical  vessel 
4  feet  deep  ?  Ans.  4000  lbs. 


Problem  II. 
To  find  the  lateral  surface  of  a  right  prism. 
4*7.  Multiply  the   length  into  the  perimeter  of  the 

BASE. 

Each  of  the  sides  of  the  prism  is  a  right  parallelogram, 
whose  area  is  the  product  of  its  length  and  breadth.  But 
the  breadth  is  one  side  of  the  base;  and  therefore,  the  sum 
of  the  breadths  is  equal  to  the  perimeter  of  the  base. 

Ex.  1.  If  the  base  of  a  right  prism  be  a  regular  hex- 
agon whose  sides  are  each  2  feet  3  inches,  and  if  the  height 
be  16  feet,  what  is  the  lateral  surface  ? 

Ans.  216  square  feet. 

If  the  areas  of  the  two  ends  be  added  to  the  lateral  sur- 
face, the  sum  will  be  the  whole  surface  of  the  prism.  And 
the  superficies  of  any  solid  bounded  by  planes,  is  evidently 
equal  to  the  areas  of  all  its  sides. 

2.  If  the  base  of  a  prism  be  an  equilateral  triangle 
whose  perimeter  is  6  feet,  and  if  the  height  be  17  feet,  what 
is  the  surface  ? 

The  area  of  the  triangle  is     1.732.  {Art.  11.) 
And  the  whole  surface  is  105.464. 


MENSURATION    OF   SOLIDS.  49 

Problem  III. 
To  find  the  soLiDirr  of  a  pyramid. 
48.  Multiply    the    area    of  the   base  into  \  of  the 

HEIOUT. 

The  solidity  of  a  prism  is  equal  to  the  product  of  the 
area  of  the  base  into  the  height.  (Art.  43.)  And  a  pyramid 
is  -^  of  a  prism  of  the  same  base  and  altitude.  (Sup.  Euc. 
15,  3.  Cor.  1.)*  Therefore  the  solidity  of  a  pyramid 
whether  right  or  oblique,  is  equal  to  the  product  of  the  base 
into  \  of  the  perpendicular  height. 

Ex.  1.  What  is  the  solidity  of  a  triangular  pyramid, 
whose  height  is  60,  and  each  side  of  whose  base  is  4  ? 

The  area  of  the  base  is      6.928 
And  the  solidity  is  138.56. 

2.  Let  ABC  be  one  side  of  an  oblique 
pyramid  whose  base  is  6  feet  square ; 
let  BC  be  20  feet,  and  make  an  angle 
of  70  degrees  with  the  plane  of  the 
base  ;  and  let  CP  be  perpendicular  to  this 
plane.  What  is  the  solidity  of  the  pyr- 
amid ? 

In  the  right  angled  triangle  BCP,  (Trig.  134.) 

R  :  BC  :  :  sinB  ::  PC  =  18.'79. 

And  the  solidity  of  the  pyramid  is  225.48  feet. 

3.  What  is  the  solidity  of  a  pyramid  whose  perpendicular 
height  is  72,  and  the  sides  of  whose  base  are  67,  54,  and 
40?  Ans.  25920. 

*  Thomson's  Legendre,  15  and  18.  7. 


44  MENSURATION    OP   SOLIDS. 


Problem  IV. 
To  find  the  lateral  surface  of  a  regular  pyramid. 

49.  Multiply  half  the  slant-height  into  the  perim- 
eter OF  the  base. 

Let  the  triangle  ABC  be  one  of 
the  sides  of  a  regular  pyramid.  As 
the  sides  AC  and  BC  are  equal,  the 
angles  A  and  B  are  equal.  Therefore 
a  line  drawn  from  the  vertex  C  to  the 
middle  of  AB  is  'perpendicular  to  AB. 
The  area  of  the  triangle  is  equal  to 
the  product  of  half  this  perpendicular  into  AB.  (Art.  8.) 
The  perimeter  of  the  base  is  the  sum  of  its  sides,  each  of 
which  is  equal  to  AB.  And  the  areas  of  all  the  equal  tri- 
angles which  constitute  the  lateral  surface  of  the  pyramid, 
are  together  equal  to  the  product  of  the  perimeter  into  half 
the  slant-height  CP. 

The  slant-height  is  the  hypothenuse  of  a  right  angled  tri- 
angle, whose  legs  are  the  axis  of  the  pyramid,  and  the  dis- 
tance from  the  centre  of  the  base  to  the  middle  of  one  of  the 
sides.     See  Def.  10. 

Ex.  1.  What  is  the  lateral  surface  of  a  regular  hexagonal 
pyramid,  whose  axis  is  20  feet,  and  the  sides  of  whose  base 
are  each  8  feet  ? 

The  square  of  the  distance  from  the  centre  of  the  base  to 
one  of  the  sides.  (Art.  16.) =48. 

The  slant-height  (Euc.  47.  l.)*=:V48-f  (20)'=21.16 

And  the  lateral  surface==21.16X 4X6=507.84  sq.  feet. 

2.  What  is  the  whole  surface  of  a  regular  triangular  pyr^ 


*  Thomson's  Legendre,  11.  4. 


ITION    OP   SOLIDS. 


4ft. 


amid  whose  axis  is  8,  and  the  sides  of  whose  base  are  each 

20.78? 

The  lateral  surface  is  312 

Tlie  area  of  the  base  is  18  Y 

And  the  whole  surface  is  499 

3.  What  is  the  lateral  surface  of  a  regular  pyramid 
whose  axis  is  12  feet,  and  whose  base  is  18  feet  square  ? 

Ans.  540  square  feet. 

The  lateral  surface  of  an  obliqiie  pyramid  may  be  found, 
by  takmg  the  sum  of  the  areas  of  the  unequal  triangles 
which  form  its  sides. 


Problem  V. 
To  Jind  the  soLiDrrr  of  a  fbustttm  of  a  pyramid. 
50.  Add  together  the  areas  of   the   two  ends,  and 

THE  square  root  OF  THE  PRODUCT  OF  THESE  AREAS  ;  AND 
MULTIPLY  THE  SUM  BY  -^  OF  THE  PERPENDICULAR  HEIGHT 
OF   THE    SOLID. 

Let  CDGL  be  a  vertical 
section,  through  the  middle 
of  a  frustum  of  a  right  pyr- 
amid CDV,  whose  base  is  a 
square. 

Let  CD=a,  LG=6,  RN=A. 

By  similar  triangles, 
LG  :  CD  ::RV  :  NV 

Subtracting  the  antecedents,  (Alg.  349.) 
LG  :  CD— LG  : :  RV  :  NV— RV=RN. 
RNxLG      hb 


l/  * 

G 

/.■ 

\ 

C                         ] 

H       i 

^           D 

Therefore  RV: 


CD— LG""a— 6 


46  MENSURATION    OF   SOLIDS. 

The  square  of  CD  is  the  base 
of  the  pyramid  CDV ; 

And  the  square  of  LG  is 
the  base  of  the  small  pyr- 
amid LGV. 


V 

i 


Therefore,  the  solidity  of 
the  larger  pyramid  (Art. 
48)  is 


p 


\       a — 0/      3a — 36 

And  the  solidity  of  the  smaller  pyramid  is  equal  to 


LG^XiRV=2''X 


3a— 36     3a— 36 


If  the  smaller  pyramid  be  taken  from  the  larger,  there 
will  remain  the  frustum  CDLG,  whose  solidity  is  equal  to 

^t::|^';=x^X^^=iAxK+a6+6^)  (Alg.l94.  a.) 
oa — do  a — 0 

Or,  because  Va'6''=a6.   (Alg.  210.  a.) 

iAX(a^+6'+Va^F) 

Here  A,  the  height  of  the  frustum,  is  multiplied  into  a' 
and  W,  the  areas  of  the  two  ends,  and  into  Va"6'*  the  square 
root  of  the  products  of  these  areas. 

In  this  demonstration  the  pyramid  is  supposed  to  be 
square.  But  the  rule  is  equally  applicable  to  a  pyramid  of 
any  other  form.  For  the  solid  contents  of  pyramids  are 
equal,  when  they  have  equal  heights  and  bases,  whatever  be 
the  Jiffure  of  their  bases.  (Sup.  Euc.  14.  3.)*     And  the  sec- 

*  Thomson's  Legendre,  14,  7. 


MENSURATION    OF   SOLIDS.  W 

tions  parallel  to  the  bases,  and  at  equal  distances,  are  equal 
to  one  another.  (Sup.  Euc.  12.  3.  Cor.  2.)* 

Ex.  1.  If  one  end  of  the  frustum  of  a  pyramid  be  9  feet 
square,  the  other  end  6  feet  square,  and  the  height  36  feet, 
what  is  the  sohdity  ? 

The  areas  of  the  two  ends  are  81  and  36. 
The  square  root  of  their  product  is  54. 
And  the  solidity  of  the  frustum=(81+36+64)xl2=2062. 

2.  If  the  height  of  a  frustum  of  {f  pyramid  be  24,  and 
the  areas  of  the  two  ends  441  and  121 ;  what  is  the  solid- 
ity? Ans.  6344. 

8.  If  the  height  of  a  frustum  of  a  hexagonal  pyramid  be 
48,  each  side  of  one  end  26,  and  each  side  of  the  other  end 
16  ;  what  is  the  solidity  ?  Ans.  56034. 

Problem  VI. 

To  find  the   lateral  surface  of  a  frustum  of  a  regular 
pyramid, 

51.  Multiply  half  Ihe  slant-heioht  by  t^k  sum  of 
THE  perimeters  of  the  two  ends. 

Each  side  of  a  frustum  of  a  regular  pyramid  is  a  trapezoid, 
as  ABCD.  The  slant-height  HP,  (Def. 
11.)  though  it  is  oblique  to  the  base  of 
the  soHd,  is  perpendicular  to  the  line  AB. 
The  area  of  the  trapezoid  is  equal  to  the 
product  of  half  this  perpendicular  into 
the  sum  of  the  parallel  sides  AB  and  DC. 
(Art.  12.)  Therefore  the  area  of  all  the 
equal  trapezoids  which  form  tho  lateral 
surface  of  the  frustum,  is  equal  to  the 


♦  Thomson's  Legendre,  13,  7,  Cor, 


48 


MENSURATION    OF    SOLIDS. 


product  of  half  the  slant-height  into  the  sum  of  the  peri- 
meters of  the  ends. 

Ex.  If  the  slant-height  of  a  frustum  of  a  regular  octag- 
onal pyramid  be  42  feet,  the  sides  of  one  end  5  feet  each,  and 
the  sides  of  the  other  end  3  feet  each ;  what  is  the  lateral 
surface?  Ans.  1344  square  feet. 

62.  If  the  slant-height  be  not  given,  it  may  be  obtained 
from      the     perpendicular 


height  and  the  dimensions 
of  the  two  ends.  Let  GD 
be  the  slant-height  of  the 
frustum  CDGL,  RN  or  GP 
the  perpendicular  height, 
ND  and  RG  the  radii  of  the 
circles  inscribed  in  the  pe- 
rimeters of  the  two  ends.  ^ 
Then.  PD  is  the  difference  of  the  two  radii : 


17 


And  the  slant-height  GD  =  v(GP'-fPD=^). 


Ex.  If  the  perpendicular  height 
of  a  frustum  of  a  regular  hexagonal 
pyramid  be  24,  the  sides  of  one  end 
13  each,  and  the  sides  of  the  other 
end  8  each ;  what  is  the  whole  sur- 
face? 


V(BC'^— BP')  =  CP,  that  is,  V(l 3'— 6^5^=11.258 
And     V8^— 4=         =   6.928 
The  difference  of  the  two  radii  is,  therefore        4.33 


Theslant-height=V(24»+4.33')=24.3875. 
TJie  lateral  surface  is  1536.4 

And  the  whole  surface,  2 141. 7 5. 


USV8URATI0N    OF   SOLIDS.  4i 

The  height  of  the  whole  pyramid  may  be  calculated  from 
the  dimensions  of  the  frustum.  Let  VN  (Fig.  17.)  be  the 
height  of  the  pyramid,  RN  or  GP  the  height  of  the  frus- 
tum, ND  and  RG  the  radii  of  the  circles  inscribed  in  the 
perimeters  of  the  ends  of  the  frustum. 

Then,  in  the  similar  triangles  GPD  and  VND, 

DP  :  GP  : :  DN  ;  VN. 

The  height  of  the  frustum  subtracted  from  VK,  gives  VR 
the  height  of  the  small  pyramid  VLG.  The  solidity  and 
lateral  surface  of  the  frustum  may  then  be  found,  by  sub- 
tracting from  the  whole  pyramid,  the  part  which  is  above 
the  cutting  plane.  This  method  may  serve  to  verify  the  cal- 
culations which  are  made  by  the  rules  in  Arts.  50  and  51. 

Ex.  If  one  end  of  the  frustum  CDGL  (Fig.  17.)  be  90  feet 
square,  the  other  end  60  feet  square,  and  the  height  RN  36 
feet ;  what  is  the  height  of  the  whole  pyramid  VCD  :  and 
what  arc  the  solidity  and  lateral  surface  of  the  frustum  ? 

DP=DN—GR=45— 30=15.       And,  GP=RN=36. 

Then,  15  :  36  : :  45  :  108 =VN,  the  height  of  the  whole 
pyramid. 

And,  108— 36=72 =VR,  the  height  of  the  part  VLG. 

The  solidity  of  the  large  pyramid  is  291600  (Art.  48.) 
of  the  small  pyramid       86400 

of  the  frustum  CDGL    205200 


The  lateral  surface  of  the  large  pyramid  is    21060  (Art.  49.) 
of  the  small  pyramid         9360 

of  the  frustum  11700 


60 


MENSURATION    OF   SOLIDS. 


Problem  VII. 
To  find  the  solidity  of  a  wedge. 
64.    Add    the    length    of    the    edge   to    twice    the 


product  of  the  height  of  the  wedge  and  the  breadth 
of  the  base. 

Let    L  =  AB    the 

length  of  the  base. 
Let  Z=GH  the  length 

of  the  edge. 
Let  6=BC  the  breadth 

of  the  base. 
Let  A=PG  the  height 

of  the  wedge. 
Then,  L 


A  M 

Z=AB— GH=AM 


If  the  length  of  the  base  and  the  edge  be  equal,  as  BM 
and  GH,  the  wedge  MBHG  is  ha,lf  a  parallelepiped  of  the 
same  base  and  height.  And  the  soHdity  (Art.  43.)  is  equal 
to  half  the  product  of  the  height,  into  the  length  and  breadth 
of  the  base  ;  that  is  i  &AZ. 

If  the  length  of  the  base  be  greater  than  that  of  the  edge, 
as  ABGH ;  let  a  section  be  made  by  the  plane  GMN,  par- 
allel to  HBC.  This  will  divide  the  whole  wedge  into  two 
parts  MBH<T  and  AMG.  The  latter  is  a  pyramid,  whose 
solidity  (Art.  48.)  is  i  hhx{L—l) 

The  solidity  of  the  parts  together,  is,  therefore, 

i6A?+,i&AX  (L— Z)=iM3Z+i&A2L— i6A2Z=i6A  X  (2L+?) 

If  the  length  of  the  base  be  less  than  that  of  the  edge,  it 
is  evident  that  the  pyi-amid  is  to  be  subtracted  from  half  the 
parallelepiped,  which  is  equal  in  height  and  breadth  to  the 
wedge,  and  equal  in  length  to  the  edge. 


MENSURATION    OF   SOLIDS.  51 

The  solidity  of  the  wedge  is,  therefore, 

Ex.  1.  If  the  base  of  a  wedge  be  35  by  15,  the  edge  55, 
and  the  perpendicular  height  12.4  ;  what  is  the  solidity  ? 

Ans.  (70  +  55)X^^^^^  =  3875. 

2.  If  the  base  of  a  wedge  be  27  by  8,  the  edge  30,  and 
the  perpendicular  height  42  ;  what  is  the  solidity? 

Ans.  5040. 

Problem  VIII. 
To  find  the  solidity  of  a  rectangular  prismoid. 

55.  To  the  areas  of  the  two  ends,  add  four  times 
the  area  of  a  parallel  section  equally  distant  from 
the  ends,  and  multiply  the  sum  by  ^  of  the  height. 

Let   L    and    B    be   the    length   and 

breadth  of  one  end, 
Let  I  and  h  be  the  length  and  breadth 

of  the  other  end. 
Let   M  and    m   be  the    length    and 

breadth  of  the  section  in  the  middle. 
And  h   be  the   height   of   the    pris- 

inoid. 

The  solid  may  be  divided  into  two  wedges  whose  bases  are 
the  ends  of  the  prismoid,  and  whose  edges  are  L  and  /.  The 
solidity  of  the  whole,  by  the  preceding  article  is, 

ii?Ax(2L+/)+i*AX(2/+L)=-iA(2BL-f-B^+2W+6L) 

As  M  is  equally  distant  from  L  and  I, 

2M=«L4-/,2/w-»B+J,and4Mm— {L+/)(B+2')=BL+B;+ 

[bh+lb. 


52  MENSURATION    OF   SOLIDS. 

Substituting  4  Mm  for  its  value,  in  the  preceding  ex- 
pression for  the  sohdity,  we  have 

ih(BL+hl+AMm) 

That  is,  the  soHdity  of  the  prismoid  is  equal  to  -J-  of  the 
height,  multiplied  into  the  areas  of  the  two  ends,  and  4 
times  the  area  of  the  section  in  the  middle. 

This  rule  may  be  applied  to  prismoids  of  other  forms. 
For,  whatever  be  the  figure  of  the  two  ends,  there  may  be 
drawn  in  each,  such  a  number  of  small  rectangles,  that  the 
sum  of  them  shall  differ  less,  than  by  any  given  quantity, 
from  the  figure  in  which  they  are  contained.  And  the  solids 
between  these  rectangles  will  be  rectangular  prismoids. 

Ex.  1.  If  one  end  of  a  rectangular  prismoid  be  44  feet  by 
23,  the  other  end  36  by  21,  and  the  perpendicular  heiglit 
72  ;  what  is  the  solidity  ? 

The  area  of  the  larger  end         =44X23  =  1012 
of  the  smaller  end       =36X21=   156 
of  the  middle  section  =40X22=   880 
And  the  solidity=  (101 2+756 +4  X  880)  X 12  =  63456  feet. 

2.  What  is  the  solidity  of  a  stick  of  hewn  timber,  whose 
ends  are  30  inches  by  27,  and  24  by  18,  and  whose  length 
is  48  feet  ?  Ans.  204  feet. 

Other  solids  not  treated  of  in  this  section,  if  they  be 
bounded  by  plane  surfaces,  may  be  measured  by  supposing 
them  to  be  divided  into  prisms,  pyramids,  and  wedges.  And, 
indeed,  every  such  solid  may  be  considered  as  made  up  of 
triangular  pyramids. 


1. 

The  Tetraedron, 

2. 

The  Ilexaedron  or  cube, 

whose 
sides  are 

3. 
4. 

The  Octaedron, 
The  Dodecaedran, 

5. 

The  Icosaedron, 

MENSURATION    OP   REGULAR   SOLIDS.  53 


THE   FIVE    REGULAR   SOLIDS. 

66.  A  SOLID  IS  SAID  TO  BE  REGULAR,  WHEN  ALL  ITS 
SOLID  ANGLES  ARE  EQUAL,  AND  ALL  ITS  6IDKS  ARE  EQUAL 
AND    REGULAR    POLYGONS. 

The  following  figures  are  of  this  description ; 

four  triangles ; 
six  squares ; 
eight  triangles ; 
twelve  pentagons; 
twenty  triangles.* 

Besides  these  five  there  can  be  no  other  regular  solids. 
The  only  plane  figures  which  can  form  such  solids,  are  tri- 
angles, squares,  and  pentagons.  For  the  plane  angles  which 
contain  any  sohd  angle,  are  together  less  than  four  right  an- 
gles or  360°.  (Sup.  Euc.  21,  2.)  And  the  least  number 
which  can  form  a  solid  angle  is  three.  (Sup.  Euc.  Def.  8,  2.) 
If  they  are  angles  of  equilateral  triangles,  each  is  60°.  The 
sum  of  three  of  them  is  180°,  of  four  240°,  oijlve  300°,  and 
of  six  360°.  The  latter  number  is  too  great  for  a  solid  angle. 

The  angles  of  squares  are  90°  each.  The  sum  of  three  of 
these  is  270°,  of  four  300°,  and  of  any  other  greater  number, 
still  more. 

The  angles  of  regular  pentaf/ons  are  108°  each.  The  sum 
of  three  of  them  is  324°  ;  of  four,  or  any  other  greater  num- 
ber, more  than  360°.  The  angles  of  all  other  regular  poly- 
gons are  still  greater. 

In  a  regular  solid,  then,  each  solid  angle  must  be  con- 
tained by  three,  four,  or  five  equilateral  triangles,  by  three 
squares,  or  by  three  regular  pentagons. 

*  For  the  geometrical  construction  of  these  solids,  see  Legendre's 
Geometry ;  Appendix  to  Books  VI.  aojl  VII.,  or  Thomson's  Legendre, 
p.  214. 


54  MENSURATION    OF   REGULAR    SOLIDS. 

67.  As  the  sides  of  a  regular  solid  are  similar  and  equal, 
and  the  angles  are  also  alike ;  it  is  evident  that  the  sides 
are  all  equally  distant  from  a  central  point  in  the  solid.  If 
then,  planes  be  supposed  to  proceed  from  the  several  edges 
to  the  centre,  they  will  divide  the  solid  into  as  many  equal 
'pyramids,  as  it  has  sides.  The  base  of  each  pyramid  will  be 
one  of  the  sides ;  their  common  vertex  will  be  the  central 
point;  and  their  height  will  be  a  perpendicular  from  the 
centre  to  one  of  the  sides. 


Problem  IX. 

To  find  the  surface  of  a  regular  solid. 

68.  Multiply  the  area  of  one  of  the  sides  by  the 
number  of  sides. 

Or, 

Multiply  the  square  of  one  of  the  edges,  by  the 
surface  of  a  similar  solid  whose  edges  are  1. 

As  all  the  sides  are  equal,  it  is  evident  that  the  area  of 
one  of  them,  multiplied  by  the  number  of  sides,  will  give  the 
area  of  the  whole. 

Or,  if  a  table  is  prepared,  containing  the  surfaces  of  the 
several  regular  solids  whose  linear  edges  are  unity  ;  this  may- 
be used  for  other  regular  solids,  upon  the  principle,  that  the 
areas  of  similar  polygons  are  as  the  squares  of  their  homolo- 
gous sides.  (Euc.  20.  6.)*  Such  a  table  is  easily  formed,  by 
multiplying  the  area  of  one  of  the  sides,  as  given  in  Art.  17, 
by  the  number  of  sides.  Thus,  the  area  of  an  equilateral 
triangle  whose  side  is  1,  is  0.4330127.  Therefore,  the  siu:- 
face 

*  Thomson's  Legendre,  27.  4. 


MEN  SI  RATION    OP   REGULAR    SOLIDS.  5ff 

Of  aregular  tetraedron  =.433012'7X4  =1.7320508. 
Of  a  regular  octaedron  =.4330127x8  =3.4641016. 
Of  a  regular  icosaedron  =.4330127x20=8.0602540. 

See  the  table  in  the  following  article. 

Ex.  1 .  What  is  the  surface  of  a  regular  dodecaedron  whose 
edges  are  each  25  inches  ? 

The  area  of  one  of  the  sides  is  1075.3 
And  the  surface  of  the  whole  solid  =1075.3X12=12903.6. 

2.  What  is  the  surface  of  a  regular  icosaedron  whose 
edges  are  each  102?  Ans.  90101.3. 

Problem  X. 
.    7b  ^rw? /^€  solidity  o/*  a  REGULAR  solid. 
59.  Multiply  the  surface  by  ^  of  the  perpendicular 

DISTANCE  from  THE  CENTRE  TO  ONE  OF  THE  SIDES. 

Or, 

McLTIPLY  the  cube  of  ONE  OF  THE  EDGES,  BY  THE 
SOLIDITY    OF    A    SIMILAR   SOLID    WHOSE   EDGES    ARE    1. 

As  the  solid  is  made  up  of  a  number  of  equal  pyramids, 
whose  bases  are  the  sides,  and  whose  height  is  the  perpendic- 
ular distance  of  the  sides  from  the  centre  (Art.  57.) ;  the 
solidity  of  the  whole  must  be  equal  to  the  areas  of  all  the 
sides  midtiplied  into  ^  of  this  perpendicular.  (Art.  48.) 

If  the  contents  pf  the  several  regular  solids  whose  edges 
are  1,  be  inserted  in  a  table,  this  may  be  used  to  measure 
other  similar  .solids.  For  two  similar  regular  solids  contain 
the  same  number  of  similar  pyramids ;  and  the.se  are  to  each 
other  as  the  cubes  of  their  linear  sides  or  edges.  (Sup.  Euc. 
15.  3.  Cor.  3.)* 

•  Thomson's  Legendre,  20.  7. 


56 


MENSURATION    OF   THE    CYLINDER. 


A    TABLE    OF   REGULAR   SOLIDS    WHOSE    EDGES    ARE    1. 


Names. 

No.  of  sides. 

Surfaces.          |        Solidities.        | 

Tetraedron 

Hexaedron 

Octaedron 

Dodecaedron 

Icosaedron 

4 

6 

8 

12 

20 

1.7320508 
6.0000000 
3.4641016 
20.6457288 
8.6602540 

0.1178513 
1.0000000 
0.4714045 
7.6631189 
2.1816950 

For  the  method  of  calculating  the  last  column  of  this  table, 
see  Hutton's  Mensuration,  Part.  III.  Sec.  2. 

Ex.  What  is  the  solidity  of  a  regular  octaedron  whose 
edges  are  each  32  inches  ?  Ans.  15447  inches. 


SECTION  IV. 


THE    CYLINDER,    CONE,    AND    SPHERE. 

Art.  61.  Definition  I.  A  right  cylinder  is  a  solid  de- 
scribed by  the  revolution  of  a  rectangle  about  one  of  its 
sides.  The  ends  or  hases  are  evidently  equal  and  parallel 
circles.  And  the  axis,  which  is  a  line  passing  through  the 
middle  of  the  cylinder,  is  perpendicular  to  the  bases. 

The  ends  of  an  oblique  cylinder  are  also  equal  and  paral- 
lel circles  ;  but  they  are  not  perpendicular  to  the  axis.  The 
height  of  a  cylinder  is  the  perpendicular  distance  from  one 
base  to  the  plane  of  the  other.  In  a  right  cylinder,  it  is  the 
length  of  the  axis. 

II.  A  right  cone  is  a  solid  described  by  the  revolution  of 
a  right  angled  triangle  about  one  of  the  sides  which  contain 
the  right  angle.    The  hase  is  a  circle,  and  is  perpendicular  to 


MENSX7RATI0N    OF   TUB   CVLINDER. 


«lr 


the  axiSf  which  proceeds  from  the  middle  of  the  base  to  the 
vertex. 

The  base  of  an  oblique  cone  is  also  a  circle,  but  is  not  per- 
pendicular to  the  axis.  The  height  of  a  cone  is  the  perpen- 
dicular distiince  from  the  vertex  to  the  plane  of  the  base.  In 
a  right  cone,  it  is  the  length  of  the  axis.  The  slant-height 
of  a  right  cone  is  the  distance  from  the  vertex  to  the  circum- 
ference of  the  base. 

III.  A  frustum  of  a  cone  is  a  portion  cut  oflf  by  a  plane 
parallel  to  the  base.     The  height  of  the 

frustum  is  the  perpendicular  distance  of 
the  two  ends.  The  slant-height  of  a 
frustum  of  a  right  cone,  is  the  distance 
between  the  peripheries  of  the  two 
ends,  measured  on  the  outside  of  the 
solid ;  as  AD. 

IV.  A  sphere  or  globe  is  a  solid  which 
has  a  centre  equally  distant  from  every 

part  of  the  surface.  It  may  be  described  by  the  revolution 
of  a  semicircle  about  a  diameter.  A  radius  of  the  sphere  is 
a  line  drawn  from  the  centre  to  any  part  of  the  surface.  A 
diameter  is  a  line  passing  through  the  centre,  and  terminated 
at  both  ends  by  the  surface.  The  circumference  is  the  same 
as  the  circumference  of  a  circle  whose  plane  passes  through 
the  centre  of  the  sphere.  Such  a  circle  is  called  a  great 
circle. 

V.  A  segment  oi  a  sphere  is  a  part  cut  off  by  any  plane. 
The  height  of  the  segment  is  a  per- 
pendicular from  the  middle  of   the 
base  to  the  convex  surface,  as  LB. 

VI.  A  spherical  zone  or  frustum  is 
a  part  of  the  sphere  included  be- 
tween two  parallel  planes.  It  is 
called  the  middle  zone,  if  the  planes 
are  equally  distant  from  the  centre. 

3* 


58  MENSURATION    OF   THE    CYLINDER. 

The  height  of  a  zone  is  the  distance  of  the  two  planes,  as 
LR.* 

VII.  A  spherical  sector  is  a  solid  produced  by  a  circular 
sector,  revolving  in  tlie  same  manner 
as  the  semicircle  which  describes  the 
whole  sphere.  Thus  a  spherical  sec- 
tor is  described  by  the  circular  sec- 
tor ACP  or  GCE  revolving  on  the 
axis  CP. 

VIII.  A  solid  described  by  the 
revolution  of  any  figure  about  a  fixed 
axis,  is  called  a  solid  of  revolution. 

Problem  I. 
To  find  the  convex  surface  of  a  right  cylinder. 
62.  Multiply  the  length   into  the  circumference  of 

THE    BASE. 

If  a  right  cylinder  be  covered  with  a  thin  substance  like 
paper,  which  can  be  spread  out  into  a  plane ;  it  is  evident 
that  the  plane  will  be  a  parallelogram,  whose  length  and 
breadth  will  be  equal  to  the  length  and  circumference  of  the 
cylinder.  The  area  must,  therefore,  be  equal  to  the  length 
multiplied  into  the  circumference.  (Art.  4.) 

Ex.  1.  What  is  the  convex  surface  of  a  right  cylinder 
which  is  42  feet  long,  and  15  inches  in  diameter? 

Ans.  42X1.25X3.14159  =  164.933  sq.  feet. 

2.  What  is  the  whole  surface  of  a  right  cylinder,  which 
is  2  feet  in  diameter  and  36  feet  long  ? 

*  According  to  some  writers,  a  spherical  segment  is  either  a  solid 
which  is  cut  off  from  the  sphere  by  a  single  plane,  or  one  which  is  in- 
cluded between  two  planes :  and  a  zone  is  the  surface  of  either  of  these. 
In  this  sense,  the  term  zone  is  commonly  used  in  geography. 


MENSURATION    OF   THB    CYLINDER.  56 

The  convex  surface  is  226.1945 

The  area  of  the  two  ends  (Art.  30.)  is  6.2832 

The  whole  surface  is  232.4777 

3.  What  is  the  whole  surface  of  a  right  cylinder  whose 
axis  is  82,  and  circumference  71  ?  Ans.  6024.32. 

63.  It  will  be  observed  that  the  rules  for  the  prism  and 
pyramid  in  the  preceding  section,  are  substantially  the  same, 
as  the  rules  for  the  cylinder  and  cone  in  this.  There  may  be 
some  advantage,  however,  in  considering  the  latter  by  them- 
selves. 

In  the  base  of  a  cylinder,  there  may  be  inscribed  a  poly- 
gon, which  shall  differ  from  it  less  than  by  any  given  space. 
(Sup.  Euc.  6.  1.  Cor.)*  If  the  polygon  be  the  base  of  a 
prism,  of  the  same  height  as  the  cylinder,  the  two  solids 
may  differ  less  than  by  any  given  quantity.  In  the  same 
manner,  the  base  of  a  pyramid  may  be  a  polygon  of  so  many 
sides,  as  to  differ  less  than  by  any  given  quantity,  from  the 
base  of  a  cone  in  which  it  is  inscribed.  A  cyhnder  is  there- 
fore considered,  by  many  writers,  as  a  prisr^  of  an  infinite 
number  of  sides ;  and  a  cone,  as  a  pyramid  of  an  infinite 
number  of  sides.  (For  the  meaning  of  the  term  "  infinite," 
when  used  in  the  mathematical  sense,  see  Alg.  Sec.  XV.) 

Problem  II. 
To  find  the  solidity  of  a  cyundkr. 

64.  Multiply  the  area  op  the  base  by  the  height. 

The  solidity  ^f  a  parallelopiped  is  equal  to  the  product  of 
the  base  into  the  perpendicular  altitude.  (Art.  43.)  And  a 
parallelopiped  and  a  cylinder  which  have  equal  bases  and 
altitudes  are  equal  to  each  other.  (Sup.  Euc.  17.  3.)f 

•  Thomson's  Legendre,  9.  5.  t  ^^^^y  2*  8. 


60 


MENSURATION    OF    THE    CYLINDER. 


Ex.  1.  What  is  the  solidity  of  a  cylinder,  whose  height  is 
121,  and  diameter  45.2  ? 

Ans.  45.2'X.V854X  121  =  194156.6. 

2.  What  is  the  solidity  of  a  cylinder,  whose  height  is  424, 
and  circumference  213  ?  Ans.   1530837. 


3.  If  the  side  AC  of  an  oblique 
cylinder  be  27,  and  the  area  of  the  base 
32.61,  and  if  the  side  make  an  angle 
of  62°  44'  with  the  base,  what  is  the 
sohdity  ? 

E  :  AC  : :  sin  A  :  BC  =  24  the   per- 
pendicular height. 

And  the  soHdity  is  782.64. 


4.  The  Winchester  bushel  is  a  hollow  cylinder,  18^  inches 
in  diameter,  and  8  inches  deep.     What  is  its  capacity  ? 
The  area  of  the  base=(18. 5)' X. 7853982  =  268.8025. 
And  the  "^^apacity  is  2150.42   cubic  inches.     See  the 
table  in  Art.  42. 


Problem  III. 

To  find  the  convex  surface  of  a  right  cone. 

65.  Multiply  half  the  slant-height  into  the  cir- 
cumference OF  the  base. 

If  the  convex  surface  of  a  right  cone  be  spread  out  into  a 
plane,  it  will  evidently  form  a  sector  of  a  circle  whose  radius 
is  equal  to  the  slant-height  of  the  cone.  But  the  area  of  the 
sector  is  equal  to  the  product  of  half  the  radius  into  the 
length  of  the  arc.  (Art.  34.)  Or  if  the  cone  be  considered 
as  a  pyramid  of  an  infinite  number  of  sides,  its  lateral  sur- 


MENSURATION    OF   THE    CONE.  61 

face  is  equal  to  the  product  of  half  the  slant-height  into  the 
perimeter  of  the  base.  (Art.  49.) 

Ex.  1.  If  the  slant-height  of  a  right  cone  be  82,  and  the 
diameter  of  the  base  24,  what  is  the  convex  surface  ? 

Ans.  41X24X3.14159=8091.3  square  feet. 

2.  If  the  axis  of  a  right  cone  be  48,  and  the  diameter  of 
le  base  72,  what  is  the  whole  surface  ? 

The  slant-height  =  V(36'+48')=60.  (Euc.  47.  1.) 
The  convex  surface  is  6786 

The  area  of  the  base  4071.6 

And  the  whole^urfaco  10857.6 

3.  If  the  axis  of  a  right  cone  be  16,  and  the  circumfer- 
ence of  the  base  75.4  ;  what  is  the  whole  surface  ? 

Ans.  1206.4. 

Problem  IV. 
To  find  the  sounrrY  of  a  cone. 
66.  Multiply  the  area   op  the    base    into  -^  of  the 

HEIGHT. 

The  solidity  of  a  cylinder  is  equal  to  the  product  of  the 
base  into  the  perpendicular  height.  (Art.  64.)  And  if  a  cone 
and  a  cylinder  have  the  same  base  and  altitude,  the  cone  is 
\  of  the  cylinder.  (Sup.  Euc.  18.  3.)f  Or  if  a  cone  be  con- 
sidercd  as  a  pyramid  of  an  infinite  number  of  sides,  the  so- 
lidity is  equal  to  the  product  of  the  base  into  \  of  the  height, 
by  Art.  48. 

Ex.  1.  What  is  the  solidity  of  a  right  cone  whose  height 
is  663,  and  the  diameter  of  whose  base  is  101  ? 

Ans.  ioPx. 7854X221  =  1770622. 

*  Thomcon's  Legendre,  4. 8.  Cor. 


62  MENSURATION    OF    THE    CONE. 

2.  If  the  axis  of  an  oblique  cone  be  738,  and  make  an 
angle  of  30°  with  the  plane  of  the  base  ;  and  if  the  circum- 
ference of  the  base  be  355,  what  is  the  solidity  ? 

Ans.  1233536. 

Problem  V.. 
To  find  the  convex  surface  of  a  frustum  of  a  right  cone. 
67.    Multiply  half  the  slant-height    by  the    sum    of 

THE  peripheries  OF  THE  TWO  ENDS. 

This  is  the  rule  for  a  frustum  of  a  pyramid  ;  (Art.  51.) 
and  is  equally  applicable  to  a  frustum  of  a  cone,  if  a  cone  be 
considered  as  a  pyramid  of  an  infinite  number  of  sides. 
(Art.  63.) 

Or  thus, 

Let  the  sector  ABV  represent  the 
convex  surface  of  a  right  cone,  (Art. 
65.)  andyDCV  the  surface  of  a  portion 
of  the  cone,  cat  off  by  a  plane  parallel 
to  the  base.  Then  will  ABCD  be  tbe 
surface  of  the  frustum. 

Let  AB=a,  DC=6,  YI>=d,  AD=h.  '^ 

Then  the  area  ABV =iaX{h-\-d)=iah+iad.  (Art.  34.) 
And  the  area  'DCY=^bd. 
Subtracting  the  one  from  the  other. 

The  area  ABDC=iaA+iac/ — ^bd. 

But  d  :  d-\-h  :\h  :  a.    (Sup.  Euc.  8.  1.)*     Therefore  ^ad^ 
ibd=ibh. 

The  surface  of  the  frustum  then,  is  equal  to 
^ah+^bh.  or^hx(a-{-b) 

*  Thomson's  Legendre,  10,  5.  Cor, 


MENSURATION    OF   THE    CONE.  63 

Cor.  The  surface  of  the  frustum  is  equal  to  the  product 
of  the  slant-height  into  the  circumference  of  a  circle  which 
is  equally  distant  from  the  two  ends.  Thus,  the  surface 
ABCD  is  equal  to  the  product  of  AD  mto  MN.  For  MN  is 
equal  to  half  the  sum  of  AB  and  DC. 

Ex.  1.  What  is  the  convex  surface  of  a  frustum  of  a  right 
cone,  if  the  diameters  of  the  two  ends  be  44  and  33,  and 
the  slant-height  84  ?  ^  Ans.  10159.8. 

2.  If  the  perpendicular  height  of  a  frustum  of  a  right 
cone  be  24,  and  the  diameters  of  the  two  ends  80  and  44, 
what  is  the  whole  surface  ? 

*      Half  the  difference  of  the  diameters  is  18. 


And  V  18'+24'=30,  the  slant-height,  (Art.  52.) 
The  convex  sur&ce  of  the  frustum  is         5843 
The  sum  of  the  areas  of  the  two  ends  is  6547 

And  the  whole  surface  is  12390 


Problem  VI. 

To  find  the  solidity  of  a  frustum  of  a  cone. 

68.  Add  toosther  the  areas  of  the  two  ends,  and 
the  square  root  of  the  product  of  these  areas  ;  and 
multiply  the  sum  by  \  of  the  perpendicular  height. 

This  rule,  which  was  given  for  the  frustum  of  a  pyramid, 
(Art.  60.)  is  equally  applicable  to  the  frustum  of  a  cone ;  be- 
cause a  cone  and  a  pyramid  which  have  equal  bases  and  alti- 
tudes are  equal  to  each  other. 

Ex.  1.  What  is  the  solidity  of  a  mast  which  is  72  feet 
long,  2  feet  in  diameter  at  one  end,  and  18  inches  at  the 
other?  Ans.  174.36  cubic  feet. 


64 


MENSURATION    OF   THE   SPHERE. 


2.  What  is  tlie  capacity  of  a  conical  cistern  which  is  9 
feet  deep,  4  feet  in  diameter  at  the  bottom,  and  3  feet  at  the 
top  ?  Ans.  87.18  cubic  feet— 652.15  wine  gallons. 

3.  How  many  gallons  of  ale  can  be  put  into  a  vat  in  the 
form  of  a  conic  frustum,  if  the  larger  diameter  be  7  feet,  the 
smaller  diameter  6  feet,  and  the  depth  8  feet  ? 


Problem  VII. 

To  find  the  surface  of  a  sphere. 

69.  Multiply  the  diameter  by  the  circumference. 

Let  a  hemisphere  be  described  by  the  quadrant  CPD, 
revolving  on  the  line  CD.  Let 
AB  be  the  side  of  a  regular  poly- 
gon inscribed  in  the  circle  of 
which  DBF  is  an  arc.  Draw  AO 
and  BN  perpendicular  to  CD, 
and  BH  perpendicular  to  AO. 
Extend  AB  till  it  meets  CD  con- 
tinued. The  triangle  AOY,  re- 
volving on  OV  as  an  axis,  will 
describe  a  right  cone.  (Defin.  2.) 
AB  will  be  the  slant-height  of  a 
frustum  of  this  cone  extending 
from  AO  to  BN.  From  G  the  middle  of  AB,  draw  GM 
parallel  to  AO.  The  surface  of  the  frustum  described  by 
AB.  (Art.  67.  Cor.)  is  equal  to 

ABxc^VcGM.* 

From  the  centre  C  draw  CG,  which  will  be  perpendicular 
to  AB,  (Euc.  3.  3.)  and  the  radius  of  a  circle  inscribed  in 


*  By  drc  GM  is  meant  the  circumference  of  a  circle  the  radius  of 
which  is  GM. 


MENSURATION  OF  THE  SPHERE.  05 

the  polygon.     The  triangles  ABH  and  CGM  are  similar,  be- 
cause the  sides  are  perpendicular,  each  to  each.     Therefore, 

HB  or  ON  :  AB  : :  GM  :  GC  : :  circ  GM  :  circ  GC. 

So  that  ONxciVc  GC=ABx«>c  GM,  that  is,  the  sur- 
face of  the  frustum  is  equal  to  the  product  of  ON  the  per- 
pendicular height,  into  circ  GC,  the  perpendicular  distance 
from  the  centre  of  the  polygon  to  one  of  the  sides. 

In  the  same  manner  it  may  l>e  proved,  that  the  surfaces 
produced  by  the  revolution  of  the  lines  BD  and  AP  about 
the  axis  DC,  are  equal  to 

ND  X  circ  GC,  and  CO  X  circ  GC. 

The  surface  of  the  whole  solid,  therefore,  (Euc.  1. 2.)  is  equal  to 

CDxcirc  GC. 

The  demonstration  is  applicable  to  a  solid  produced  by 
the  revolution  of  a  polygon  of  any  number  of  sides.  But  a 
polygon  may  be  supposed  which  shall  differ  less  than  by 
any  given  quantity  from  the  circle  in  which  it  is  inscribed  ; 
(Sup.  Euc.  4.  1.)*  and  in  which  the  perpendicular  GC  shall 
differ  less  than  by  any  given  quantity  from  the  radius  of  the 
circle.  Therefore,  the  surface  of  a  hemisphere  is  equal  to 
the  product  of  its  radius  into  the  circumference  of  its  base  ; 
and  the  surface  of  a  sphere  is  equal  to  the  product  of  its 
diameter  into  its  circumference. 

Cor.  1.  From  this  demonstration  it  follows,  that  the  sur- 
face of  any  segment  or  zone  of  a  sphere  is  equal  to  the 
product  of  the  height  of  the  segment  or  zone  into  the  cir- 
cumference of  the  sphere.  The  surface  of  the  zone  pro- 
duced by  the  revolution  of  the  arc  AB  about  ON,  is  equal 
to  ON  X«rc  CP.     And  the  surface  of  the  segment  pro- 

♦  Thomson's  Legendre,  9.  5. 


66 


MENSURATION    OF   THE    SPHERE. 


duced  by  the  revolution  of  BD  about  DN  is  equal  to  DNx 
ci7'c  CP. 

Cor.  2.  The  surface  of  a  sphere  is  equal  to  four  times  tlie 
area  of  a  circle  of  the  same  diameter ;  and  therefore,  the 
convex  surface  of  a  hemisphere  is  equal  to  twice  the  area  of 
its  base.  For  the  area  of  a  circle  is  equal  to  the  product  of 
half  the  diameter  into  half  the  circumference  ;  (Art.  30.) 
that  is,  to  ^  the  product  of  the  diameter  and  circumference. 

Cor.  3.  The  surface  of  a  sphere,  or  the  convex  surface 
of  any  spherical  segment  or  zone, 
is  equal  to  that  of  the  circum- 
scribing cylinder.  A  hemis- 
phere described  by  the  revolu- 
tion of  the  arc  DBP,  is  cir- 
cumscribed by  a  cylinder  pro- 
duced by  the  revolution  of  the 
parallelogram  Dc?CP.  The  con- 
vex surface  of  the  cylinder  is 
equal  to  its  height  multiplied 
by  its  circumference.  (Art.  62.) 
And  this  is  also  the  surface  of 
the  hemisphere. 

So  the  surface  produced  by  the  revolution  of  AB  is  equal 
to  that  produced  by  the  revolution  of  ab.  And  the  surface 
produced  by  BD  is  equal  to  that  produced  by  hd. 

Ex.  1.  Considering  the  earth  as  a  sphere  7 930  miles  in 
diameter,  how  many  square  miles  are  there  on  its  surface  ? 

Ans.   197,558,500. 

2.  If  the  circumference  of  the  sun  be  2,800,000,  what  is 
his  surface?  Ans.  2,495,547,600,000  sq.  miles. 

3.  How  many  square  feet  of  lead  will  it  require,  to  cover 
a  hemispherical  dome  whose  base  is  13  feet  across  ? 

Ans.  265i-. 


UBNSITBATIOK    OF   THE   SrUERE.  67 

Problem  VIII. 
To  find  the  solidiit  of  a  sphere. 
70.  1.  Multiply  tse  cube  of  the  diameter  by  •5236. 

Or, 

2.  Multiply  the  square  of  the  diameter  by  \  of  the 
circumfekence. 

Or, 

3.  Multiply  the' surface  by  \  of  the  diameter, 

1.  A  sphere  is  two-thirds  of  its  circumscribing  cylinder. 
(Sup.  Euc.  21.  3.)*  The  height  and  diameter  of  the  cylin- 
der  are  each  equal  to  the  diameter  of  the  sphere.  The  solid- 
ity of  the  cylinder  is  equal  to  its  height  multiplied  into  the 
area  of  its  base,  (Art.  64.)  that  is  putting  D  for  the  diam- 
eter, 

DxD'X.7854     or     D'x.7854. 

And  the  solidity  of  the  sphere,  being  f  of  this,  is 
D'X.6236. 

2.  The  base  of  the  circumscribing  cylinder  is  equal  to  half 
the  circumference  multiplied  into  half  the  diameter ;  (Art. 
30.)  that  is,  if  C  be  put  for  the  circumference, 

iC  X  D  ;  and  the  solidity  is  \Q  X  D'. 

Therefore,  the  solidity  of  the  sphere  is 
iofiCxD«=D«XiC. 

8.  In  the  last  expression,  which  is  the  same  as  CxDxiD, 
•  ThomMa's  Legendre,  12.  8. 


68  MENSURATION    OF   THE    SPHERE. 

we  may  substitute  S,  the  surface,  for  CxD.  (Art.  69.)    We 
then  have  the  solidity  of  the  sphere  equal  to 

SxiD. 

Or,  the  sphere  may  be  supposed  to  be  filled  with  small 
pyramids,  standing  on  the  surface  of  the  sphere,  and  having 
their  common  vertex  in  the  centre.  The  number  of  these 
may  be  such,  that  the  difference  between  their  sum  and  the 
sphere  shall  be  less  than  any  given  quantity.  The  solidity 
of  each  pyramid  is  equal  to  the  product  of  its  base  into  -|- 
of  its  height.  (Art.  48.)  The  solidity  of  the  whole,  there- 
fore, is  equal  to  the  product  of  the  surface  of  the  sphere 
into  "i  of  its  radius,  or  ^  of  its  diameter. 

71.  The  numbers  3.14159,  .7854,  .5236,  should  be  made 
perfectly  familiar.  The  first  expresses  the  ratio  of  the 
circumference  of  a  circle  to  the  diameter;  (Art.  23.)  the 
second,  the  ratio  of  the  area  of  a  circle  to  the  square  of  the 
diameter  (Art.  30.)  ;  and  the  third,  the  ratio  of  the  solidity 
of  a  sphere  to  the  cube  of  the  diameter.  The  secon(J*is  -J- 
of  the  first,  and  the  third  is  ^  of  the  first. 
.  As  these  numbers  are  frequently  occurring  in  mathemat- 
ical investigations,  it  is  common  to  represent  the  first  of  them 
by  the  Greek  letter  n.     According  to  this  notation, 

7r=3.14159,        i7r=.7854,        i^=.5236. 

If  D=the  diameter,  and  R=:the  radius  of  any  circle  or 
sphere ; 

Then,     D  =  2R     D^=4R^     D'r=8R^ 

And  nD  )  ^,^^^       -j^^  i-D=  )  =the  area  oi-nW  )  ^^^^ 
Or,  27rR  j  ^      ^    or  ttR^  S       the  circ.  or  f^iR^  \ 

solidity  of  the  sphere. 

Ex  1.  What  is  the  solidity  of  the  earth,  if  it  be  a  sphere 
7930  miles  in  diameter  ? 

Ans.  261,107,000,000  cubic  miles. 


MENSURATION    OF   THE   SPHERE.  60 

2,  How  many  wine  gajlons  will  fill  a  hollow  sphere  4  feet 
in  diameter  ? 

Ans.  The  capacity  is  33.5104  fect=250f  gallons. 

3.  If  the  diameter  of  the  moon  be  2180  miles,  what  is  its 
sohdity?  Ans.  6,424,600,000  miles. 

72.  If  the  solidity  of  a  sphere  be  ^Iven,  the  diameter  may 
be  found  by  reversing  the  first  rule  in  the  preceding  article ; 
that  is,  dividing  hy  .5236  and  extracting  the  cube  root  of  the 
quotient, 

Ex.  1.  What  is  the  diameter  of  a  sphere  whose  solidity  is 
65.45  cubic  feet  ?  Ans.  5  feet. 

2.  What  must  be  the  diameter  of  a  globe  to  contain  16755 
poimds  of  water?  Ans.  8  feet. 

Problem  IX. 

To  find  the  convex  surface  of  a  segment  or  zone  of  a 
spJiere. 

73.  Multiply  the   height   of  the    segment   or  zone 

INTO   the    circumference    OF   THE   SPHERE. 

For  the  demonstration  of  this  rule,  see  Art.  69. 

Ex.  1.  If  the  earth  be  considered  a  perfect  sphere  7930 
miles  in  diameter,  and  if  the  polar  circle  be  23°  28'  from  the 
pole,  how  many  square  miles  are  there  in  one  of  the  frigid 
zones  ? 

If  PQOE  be  a  meridian  on  the 
earth,  ADB  one  of  the  polar  circles, 
and  P  the  pole ;  then  the  frigid  zone 
is  a  spherical  segment  described  by 
the  revolution  of  the  arc  APB  about 
PD.  The  angle  ACD  subtended  by 
the  arc  AP  is  23°  28'.  And  in  the 
rififht  angled  triangle  ACD, 


10 


MENSURATION    OF   THE   SPHERE. 

R  :  AC  : :  COS  ACD  ;  CD=3C37. 


Then,   CP~CD=3965— 363'7=328=PD    the   height  of 
the  segment. 

And  328X^930X3.14159=8]  71400  the  surface. 

2.  If  the  diameter  of  the  earth  be  7930  miles,  what  is  the 
surface  of  the  torrid  zone,  extending 
23°  28'  on  each  side  of  the  equator  ? 

If  EQ  be  the  equator,  and  GH  one 
of  the  tropics,  then  the  angle  ECG  is 
23°  28'.  And  in  the  right  angled 
triangle  GCM, 

R  :  CG  :  :  sin  ECG  :  GM=C]Sr=15'78.9  the  height  of 
half  the  zone. 

The  surface  of  the  whole  zone  is  78669700. 

3.  What  is  the  surface  of  each  of  the  temperate  zones  ? 

The  height  DN=CP— CN— PD  =  2058.1 

And  the  surface  of  the  zone  is  51273000. 

The  surface  of  the  two  temperate  zones  is  102,546,000 

of  the  two  frigid  zones  16,342,800 

of  the  torrid  zone  78,669,700 

of  the  whole  globe  197,558,500 


Problem  X. 

To  find  the  solidity  of  a  spherical  sector. 

74.    Multiply    the    spherical    surface   by    ^   of  the 
radius  of  the  sphere. 


The  spherical  sector  produced  by  the  revolution  of  ACBD 


laNSURATION    OF   THE    SPHERfi. 


a 


about  CD,  may  be  supposed  to  be  filled 

with  small  pyramids,  standing  on  the 

spherical  surface  ADB,  and  temainating 

in  the  point  C.     Their  number  may  be 

so  great,  that  the  height  of  each  shall 

differ    less   than    by  any  given    length 

from  the  radius  CD,  and  the  sum  of  their 

bases  shall  differ  less  than  by  any  given 

quantity  from  the  surface  ABD.     The 

solidity  of  each  is  equal  to  the  product  of  its  base  into  \  of 

the  radius  CD.  (Art.  48.)     Therefore,  the  solidity  of  all  of 

them,  that  is,  of  the  sector  ADBC,  is  equal  to  the  product 

of  the  spherical  surface  into  i  of  the  radius. 

Ex.  Supposing  the  earth  to  be  a 
sphere  7930  miles  in  diameter,  and 
the  polar  circle  ADB  to  be  23°  28' 
from  the  pole ;  what  is  the  solidity  of 
the  spherical  sector  ACBP  ? 

Ans.  10,799,867,000  miles. 


Problem  XI. 

To  find  the  solidity  of  a  spherical  segment. 

75,  Multiply  half  the  heioht  of  the  segment  into 
the  area  of  the  base,  and  the  cube  of  the  height 
into  .5236  ;  and  add  the  two  products. 

As  the  circular  sector  AOBC  consists  of  two  parts,  the 
segment  AOBP  and  the  triangle 
ABC ;  (Art.  35.)  so  the  spherical 
sector  produced  by  the  revolution  of 
AOC  about  OC  consbts  of  two  parts, 
the  segvient  produced  by  the  revolu- 
tion of  AOP,  and  the  cone  produced 
by  the  revolution  of  ACP.     If  then 


12 


MENSURATION    OF   THE    SPHERE. 


the  cone  be  subtracted  from  the  sec- 
tor, the  remainder  will  be  the  seg- 
ment. 

Let  CO=R,  the  radius  of  the  sphere, 
PB=r,  the  radius  of  the  base  of 

the  segment. 
PO=A,  the  height  of  the  segment, 
Then  PC=R — h,  the  axis  of  the  cone. 


The  sectors 27iRx^XiR.  (Arts.  '71,  73,  Y4.)=f;rAR«. 
The  cone=7r/^xi  (R— A) (Arts.  71,  66.)=i7rr"R — \7Tkr'' 

Subtracting  the  one  from  the  other, 

The  segment  =-|7rAR2— |;rr^R+|7rAr^ 

ButDOxPO=BO'  (Trig.  97.*)=PO^-fPB^  (Euc.  47.  1.) 

That  is,  2R7i=:A^-f  r^     So  that,  R=^'+^' 

2h 


And  R^= 


A'+7-\=     A4-f2AV='+H 


2=  /^_j:!_\  = 

V    2A    / 


W 


Substituting  then,  for  R  and  R^,  their  vahies,  and  multi- 
plying the  factors, 

The  segment=i7r7i='  ^\nhr''  -\-\-j^ — i^Ar"— iy  +\n1ir^ 
Which,  by  uniting  the  terms,  becomes 

The  first  term  here  is  ihx^r^,  half  the  height  of  the  seg- 
ment multiplied  into  the  area  of  the  base  ;  (Art.  71.)  and  the 
other  h^Xi^,  the  cube  of  the  height  multiplied  into  .5236. 


*  Euclid  31,  3,  and  8,  6.   Cor. 


MENSURATION   OF   THE   SPHERE. 


IS 


If  the  segment  be  greater  than  a  hemisphere,  as  ABD ; 
the 'cone  ABC  must  be  added  to  the  sector  ACBD. 

Let  PD=A  the  height  of  the  segment, 
Then  PC=A — R  the  axis  of  the  cone. 

The  sector  ACBD=inAR« 

The  cone=nr*Xi(A— R)=Wir' — \rtr*R 

Adding  them  together,  we  have  as  before. 

The  segment  =inhR'' — \nr'R-^inhr\ 

Cor.  The  solidity  of  a  spherical  segment  is  equal:  to  half  a 
cylinder  of  the  same  base  and  height  +  a  sphere  whose 
diameter  is  the  height  of  the  segment.  For  a  cylinder  is 
equal  to  its  height  multiplied  into  the  area  of  its  base  ;  and 
a  sphere  is  equal  to  the  cube  of  its  diameter  multiplied  by 
.5230. 


Thus,  if  Oy  be  half  Ox,  the  spher- 
ical segment  produced  by  the  revo- 
tution,  of  Oxt  is  equal  to  the  cylin- 
der produced  by  ivt/x  -f  the  sphere 
produced  by  Oyxz  ;  supposing  each 
to  revolve  on  the  line  Ox. 


Ex.  1.  If  the  height  of  a  spherical  segment  be  8  feet,  and 
the  diameter  of  its  base  25  feet ;  what  is  the  solidity  ? 

Ans.  (25)'X.V854X4+8'X.5236=2231.58  feet. 

2.  If  the  earth  be  a  sphere  7930  miles  in  diameter,  and  the 
polar  circle  23°  28'  from  the  pole,  what  is  the  solidity  of 
one  of  the  frigid  zones  ?  Ans.  1,303,000,000  miles. 

4 


I 


14 


MENSURATION    OF   THE   SPHERE. 


«  Problem  XII. 

To  find  the  SOLIDITY  of  a  sph  rlcal  zone  or  frustum, 

16.  From  the  solidity  of  the  whole  sphere,  sub- 
tract  THE   TWO    segments    ON   THE    SIDES    OF   THE   ZONE. 

Or, 

Add  together  the  squares  of  the  radii  of  the  two 
ends,  and  ^  the  square  of  their  distance  ;  and  multiply 
the  sum  by  three  times  this  distancc,  and  the  product 
BY  .5236. 

If  from  the  whole  sphere,  there 
be  taken  the  two  segments  ABP  and 
GHO,  there  will  remain  the  zone 
or  frustum  ABGH. 

Or,  the  zone  ABGH  is  equal  to 
the  difference  between  the  segments 
GHP  and  ABP. 

i  the  Jieights  of  the  two  segments 


the  radii  of  their  bases. 


DP 

GN=R  ] 
AD=r    ) 

DN=c?=H — h  the  distance  of  the  two  bases,  or  the 
height  of  the  zone. 

Then  the  larger  segment=i7rHR'+^rrH^  )  ,  .    .    ^    . 
And  the  smaller  segment=^rrAr^+-6-^^^      i 

Therefore  the  zone  ABGH=i7r  (SHR'^+H^— 3^r'— A^) 

By  the  properties  of  the  circle,  (Euc.  35,  3.) 

0]S^xH=:R^     Therefore,  (ON+H)xH=R^+H=' 

R=+H^ 


Or,  0P= 


H 


MENSURATION    OF   THE    SPHERE.  76 

In  the  same  manner,  OP=!li — 

h 

Therefore,  3Hx(r'+A*)=3Ax(R'+H*.) 

Or,  8Hr«+3lIA'— 3AR'— 3AH'=0.     (Alg.  l'/8.) 

To  reduce  the  expression  for  the  sohdity  of  the  zone  to 
the  required  form,  without  altering  its  value,  let  these  terms 
be  added  to  it :  and  it  will  become 

in(3HR'+3Hr«— 3AR'— 3Ar»+H«— 3H'A+3HA»— A3) 

Which  is  equal  to 

i^r  X  3(H— A)  X  (R'+r'+i  (H-~A)») 

Or,  as  \n  equals  .6236  (Art.  71.)  and  H — h  equals  d, 

The  zone=.6236X3rfx(R'+r2+i<i'.) 

Ex.  1.  If  the  diameter  of  one  end  of  a  spherical  zone  is 
24  feet,  the  diameter  of  the  other  end  20  feet,  and  the  dis- 
tance of  the  two  ends,  or  the  height  of  the  zone  4  feet ; 
what  is  the  solidity  ?  Ans.  1566.6  feet. 

2.  If  the  earth  be  a  sphere  7930  miles  in  diameter,  and 
the  obliquity  of  the  ecliptic  23**  i?8' ;  what  is  the  solidity  of 
one  of  the  temperate  zones  ? 

Ans.  55,390,500,000  miles. 

3.  What  is  the  solidity  of  the  torrid  zone  ? 

Ans.  147,720,000,000  miles. 

The  solidity  of  the  two  temperate  zones  is  110,781,000,000 
of  the  two  frigid  zones  2,606,000,000 

of  the  torrid  zone  147,720,000,000 

of  the  whole  globe  261,107,000,000 

4.  What  is  the  convex  surface  of  a  spherical  zone,  whose 
breadth  is  4  feet,  on  a  sphere  of  25  feet  diameter  ? 


76  MENSURATION    OF   SOLIDS. 

5.  What  is  the  solidity  of  a  spherical  segment,   whose 
teight  is  18  feet,  and  the  diameter  of  its  base  40  feet  ? 


PROMISCUOUS   EXAMPLES   OF   SOLIDS. 

Ex.  1.  How  much  water  can  be  put  into  a  cubical  vessel 
three  feet  deep,  which  has  been  previously  filled  with  cannon 
balls  of  the  same  size,  2,  4,  6,  or  9  inches  in  diameter,  regu- 
larly arranged  in  tiers,  one  directly  above  another  ? 

Ans.  96^  wine  gallons. 

2.  If  a  cone  or  pyramid,  whose  height  is  three  feet,  be 
divided  into  three  equal  portions,  by  sections  parallel  to  the 
base  ;  what  will  be  the  heights  of  the  several  parts  ? 

Ans.  24,961,  6-.488,  and  4.551  inches. 

3.  What  is  the  solidity  of  the  greatest  square  prism  which 
can  be  cut  from  a  cylindrical  stick  of  timber,  2  feet  6 
inches  in  diameter  and  56  feet  long  ?* 

Ans.  175  cubic  feet. 

4.  How  many  such  globes  as  the  earth  are  equal  in  bulk 
to  the  sun;  if  the  former  is  7930  miles  in  diameter,  and  the 
latter  890,000  ?  Ans.  1,413,678. 

*  The  common  rule  for  measuring  round  timber  is  to  multiply  the 
square  of  the  quarter-girt  by  the  length.  The  quarter-girt  is  one-fourth 
of  the  circumference.  This  method  does  not  give  the  whole  solidity.  It 
makes  an  allowance  of  about  one-fifth,  for  waste  in  hewing,  bark,  &c. 
The  solidity  of  a  cyUnder  is  equal  to  the  product  of  the  height  into  the 
area  of  the  base. 

If  C=the  circumference,  and  t=3.14159,  then  (Art.  31.) 

The  area  of  the  base^— =(—--)=:(— —-) 
4s-     \  y/47r/     V3.545/ 

If  then  the  circumference  were  divided  by  3.545,  instead  of  4,  and  the 
quotient  squared,  the  area  of  the  base  would  be  correctly  found.  See 
note  B. 


MENSURATION    OF   SOLIDS.  77 

5.  How  many  cubic  feet  of  wall  are  there  in  a  conical 
tower  66  feet  high,  if  the  diameter  of  the  base  be  20  feet 
from  outside  to  outside,  and  the  diameter  of  the  top  8  feet ; 
the  thickness  of  the  wall  being  4  feet^t  the  bottom,  and  de- 
creasing regularly,  so  as  to  be  only  two  feet  at  the  top  ? 

Ans.  Viae. 

6.  If  a  metallic  globe  filled  with  wine,  which  cost  as  much 
at  5  dollars  a  gallon,  as  the  globe  itself  at  20  cents  for  every 
square  inch  of  its  surface ;  what  is  the  diameter  of  the  globe  ? 

Ans.  55.44  inches. 

7.  If  the  circumference  of  the  earth  be  25,000  miles, 
what  must  be  the  diameter  of  a  metallic  globe,  which,  when 
drawn  into  a  wire  j^  oi  an  inch  in  diameter,  would  reach 
round  the  earth  ?  Ans.  15  feet  and  1  inch. 

8.  If  a  conical  cistern  be  3  feet  deep,  7^  feet  in  diameter 
at  the  bottom,  and  5  feet  at  the  top  ;  what  will  be  the  depth 
of  a  fluid  occupying  half  its  capacity  ? 

Ans.  14.535  inches. 

9.  If  a  globe  20  inches  in  diameter,  be  perforated  by  a 
cylinder  16  inches  in  diameter,  the  axis  of  the  latter  passing 
through  the  centre  of  the  former ;  what  part  of  the  solidity, 
and  the  surface  of  the  globe,  will  be  cut  away  by  the  cyl- 
inder ? 

Ans.  3284  inches  of  the  solidity,  and  502,655  of  the  surface. 

10.  What  is  the  solidity  of  the  greatest  cube  which  can 
be  cut  from  a  sphere  three  feet  in  diameter  ? 

Ans.  6f  feet. 

11.  What  i3  the  solidity  of  a  conic  fhistum,  the  altitude 
of  which  is  36  feet,  the  greater  diameter  16,  and  the  lesser 
diameter  8  f 

12.  What  is  the  solidity  of  a  spherical  segment  4  feet 
high,  cut  from  a  sphere  16  feet  in  diameter  ? 


VS  ISOPERIMETRY. 


SECTION    V. 


ISOPERIMETRY. 


Art.  77.  It  is  often  necessary  to  compare  a  number  of 
different  figures  or  solids,  for  the  purpose  of  ascertaining 
which  has  the  greatest  area,  within  a  given  perimeter,  or  the 
greatest  capacity  under  a  given  surface.  We  may  have  oc- 
casion to  determine,  for  instance,  what  must  be  the  form  of 
a  fort,  to  contain  a  given  number  of  troops,  with  the  least 
extent  of  wall ;  or  what  the  shape  of  a  metallic  pipe  to  con- 
vey a  given  portion  of  water,  or  of  a  cistern  to  hold  a  given 
quantity  of  liquor,  with  the  least  expense  of  materials. 

78.  Figures  which  have  equal  perimeters  are  called  Iso- 
perimeters.  When  a  quantity  is  greater  than  any  other  of  the 
same  class,  it  is  called  a  maximum.  A  multitude  of  straight 
lines,  of  different  lengths,  may  be  drawn  within  a  circle. 
But  among  them  all,  the  diameter  is  a  maximum.  Of  all 
sines  of  angles,  which  can  be  drawn  in  a  circle,  the  sine  of 
90°  is  a  maximum. 

When  a  quantity  is  less  than  any  other  of  the  same  class, 
it  is  called  a  minimum.  Thus,  of  all  straight  lines  drawn 
from  a  given  point  to  a  given  straight  line,  that  which  is  per- 
pendicular to  the  given  line  is  a  minimum.  Of  all  straight 
lines  drawn  from  a  given  point  in  a  circle,  to  the  circumfer- 
ence, the  maximum  and  the  minimum  are  the  two  parts  of 
the  diameter  which  pass  through  that  point.  (Euc.  7,  3.) 

In  isoperimetry,  the  object  is  to  determine,  on  the  one 
hand,  in  what  cases  the  area  is  a  maximum,  within  a  given 
perimeter ;  or  the  capacity  a  maximum,  within  a  given  sur- 
face :  and  on  the  other  hand,  in  what  cases  the  perimeter  is 


ISOPERIMETRY.  Vd 

a  minimum  for  a  given  area,  or  the  surface  a  minimumf  for  a 
given  capacity.  ^ 

Proposition  I. 

79.  An  Isosceles  Triangle  has  a  greater  area  than  any 
scalene  triangle,  of  equal  base  and  perimeter. 

If  ABC  be  an  isosceles  trian- 
gle whose  equal  sides  are  AC  and 
i)C  ;  and  if  ABC  be  a  scalene  tri- 
angle on  the  same  base  AB,  and 
having  AC'-f-BC'  =  AC+BC; 
ilien  the  area  of  ABC  is  greater 
than  that  of  ABC. 

Let   perpendiculars   be    raised 
from  each  end  of  the  base,  extend 
AC  to  D,  make  CD'  equal  to  AC,  join  BD,  and  draw  CH 
and  CH'  parallel  to  AB. 

As  the  angle  CAB=ABC,  (Euc.  5,  1.)  and  ABD  is  a  right 

angle,  ABC-f  CBD=CAB+CDB  =  ABC+CDB.  Therefore 

CBD  =  CDB,  so  that  CD=CB  ;  and  by  constniction,  CD'= 

AC.    The  perpendiculars  of  the  equal  right  angled  triangles 

CHD  and  CHB  are  equal;   therefore,  BH=iBD.     In  the 

ime  manner,  AH'=iAD'.     The  line  AD=AC+BC=AC 

HBC'=D'C+BC'.     But  D'C+BOBD'.    (Euc.  20,  1.) 

rhcrefore,  AD>BD' ;  BD>AD',  (Euc.  47,  1.)  and  i  BD> 

i  AD'.     But  iBD,  or  BH,  is  the  height  of  the  isosceles  tri- 

ngle  ;  (Art.  1.)  and  |AD'  or  AH',  the  height  of  the  scalene 

i  iii'^'le  ;  and  the  areas  of  two  triangles  which  have  the  same 

ire  as  their  heights.  (Art.  8.)     Therefore  the  area  of 

\ijC  is  greater  than  that  of  ABC.     Among  all  triangles, 

;  lien,  of  a  given  perimeter,  and  upon  a  given  base,  the  isos 

eles  triangle  is  a  maximum. 

Cor.  The  isosceles  triangle  has  a  less  perimeter  than  any 

■calene  triangle  of  the  same  base  and  area.     The  triangle 


80 


ISOPERIMETRY. 


ABC  being  less  than  ABC,  it  is  evident  the  perimeter  of  the 
former  must  be  enlarged,  to  make  its  area  equal  to  the  area 
of  the  latter. 


Proposition  II. 

80.  A  triangle  in  which  two  given  sides  make  a  right 
ANGLE,  has  a  greater  area  than  any  triangle  in  which  the  same 
sides  maJce  an  oblique  angle. 

If  BC,  BC  and  BC"  be  equal, 
and  if  BC  be  perpendicular  to 
AB  ;  then  the  ric^ht  anfyled  trian- 
gle  ABC,  has  a  greater  area  than 
the  acute  angled  triangle  2VBC',  or 
the  oblique  angled  triangle  AB  C". 

Let  P'C  and  PC"  be  perpen- 
dicular to  AP.  Then,  as  the 
three  triangles  have  the  same  base  AB,  their  areas  are  as 
their  heights ;  that  is,  as  the  perpendiculars  BC,  P'C,  and 
PC".  But  BC  is  equal  to  BC,  and  therefore  greater  than 
P'C.  (Euc.  47.  1.)  BC  is  also  equal  to  BC",  and  therefore 
greater  than  PC". 


Proposition  III. 

81.  If  all  the  sides  except  one  of  a  polygon  be  given, 
the  area  will  be  the  greatest,  when  the  given  sides  are  so  dis- 
2)osed  that  the  figure  may  he  inscribed  in  a  semicircle,  of 
which  the  undetermined  side  is  the  diameter. 


If  the  sides  AB,  BC,  CD,  DE, 
be  given,  and  if  their  position 
be  such  that  the  area,  included 
between  these  and  another  side 
whose  length  is  not  determined, 
is  a  nmximum  ;  the  figure  may 


I 


ISOPERIMETRY.  81 

be  inscribed  in  a  semicircle,  of  which  the  undetermined  side 
AE  is  the  diameter. 

Draw  the  lines  AD,  AC,  EB,  EC.  By  varying  the  angle 
at  D,  the  triangle  ADE  may  be  enlarged  or  diminished,  with- 
out affecting  the  area  of  the  other  parts  of  the  figure.  The 
whole  area,  therefore,  cannot  be  a  maximum,  unless  this  tri- 
angle be  a  maximum,  while  the  sides  AD  and  ED  are  given. 
But  if  the  triangle  ADE  be  a  maximum,  under  these  con- 
ditions, the  angle  ADE  is  a  right  angle ;  (Art.  80.)  and 
therefore  the  point  D  is  in  the  circumference  of  a  circle,  of 
^vhich  AE  is  the  diameter.  (Euc.  31,3.)  In  the  same  man- 
ner it  may  be  proved,  that  the  angles  ACE  and  ABE  are 
right  angles,  and  therefore  that  the  points  C  and  B  are  in 
tlie  circumference  of  the  same  circle. 

The  term  polygon  is  used  in  this  section  to  include  trian- 
fjles,  and  four-sided  figures,  as  well  as  other  right-lined 
figures. 

82.  The  area  of  a  polygon,  inscribed  in  a  semicircle,  in 
the  manner  stated  above,  will  not  be  altered  by  varying  the 
order  of  the  given  sides. 

The  sides  AB,  BC,  CD,  DE,  are  the  chords  of  so 
many  arcs.  The  sum  of  these  arcs,  in  whatever  order 
they  are  arranged,  will  evidently  be  equal  to  the  semicircum- 
ference.  Audi  the  segments  between  the  given  sides  and  the 
arcs  will  be  the  same  in  whatever  part  of  the  circle  they  are 
-ituated.  But  the  area  of  the  polygon  is  equal  to  the  area 
f  the  semicircle,  diminished  by  the  sum  of  these  segments. 

83.  If  a  polygon,  of  which  all  the  sides  except  one  are 
given,  be  inscribed  in  a  semicircle  whose  diameter  is  the  un- 
determined side  ;  a  polygon  having  the  same  given  sides, 
cannot  be  inscribed  in  any  other  semicircle  which  is  either 
greater  or  less  than  this,  and  whose  diameter  is  the  undeter- 
mined side. 

The  given  sides  AB,  BC,  CD,  DE,  are  the  chords  of  arcs 
whose  sum  is  180  degrees.     But  in  a  larger  circle,  each 

4* 


82 


ISOPERIMETRY. 


would  be  the  chord  of  a  less  number  of  degrees,  and  there- 
fore the  sum  of  the  .arcs  would  be  less  than  180° :  and  in  a 
smaller  circle,  each  would  be  the  chord  of  a  greater  number 
of  degrees,  and  the  sum  of  the  arcs  would  be  greater  than 
180°. 


Proposition  IV. 

84.  A  polygon  inscribed  in  a  circle  has  a  greater  area, 
than  any  polygon  of  equal  perimeter,  and  the  same  number  of 
sides,  which  cannot  he  inscribed  in  a  circle. 

If  in  the  circle  ACHF,   (Fig.  30.)  there  be  inscribed  a 
c 


polygon  ABCDEFG ;  and  if  another  polygon  ahcdefg  (Fig. 
31.)  be  formed  of  sides  which  are  the  same  in  number  and 
length,  but  which  are  so  disposed,  that  the  figure  cannot  be 
inscribed  in  a  circle;  the  area  of  the  former  polygon  is 
greater  than  that  of  the  latter. 

Draw  the  diameter  AH,  and  the  chords  DH  and  EH. 
Upon  de  make  the  triangle  deh  equal  and  similar  to  DEH, 
and  join  ah.  The  line  ah  divides  the  figure  abcdhefg  into  two 
parts,  of  which  one  at  least  cannot,  by  supposition,  be  in- 
scribed in  a  semicircle  of  which  the  diameter  is  AH,  nor  in 
any  other  semicircle  of  which  the  diameter  is  the  undeter- 
mined side.  (Art.  83.)  It  is  therefore  less  than  the  corres- 
ponding part  of  the  figure  ABCDHEFG.  (Art.  81.)  And 
the  other  part  of  abcdhefg  is  not  greater  than  the  correspond- 


ISOPERIMETRY.  83 

ing  part  of  ABCDHEFiS.  Therefore,  the  whole  figure 
ABCDIIEFG  is  greater  than  the  whole  figure  ahcdhefy.  If 
from  these  there  be  taken  the  equal  triangles  DEU  and  dch, 
there  will  remain  the  polygon  ABCDEFG  greater  than  the 
polygon  ahcdcfg. 

85.  A  polygon  of  which  all  the  sides  are  given  in  num- 
ber and  length,  cannot  be  inscribed  in  circles  of  different 
diameters.  (Art.  83.)  And  the  area  of  the  polygon  will  not 
be  altered  by  changing  the  order  of  the  sides.  (Art.  82.) 

Proposition  V. 

86.  When  a  polygon  has  a  greater  area  than  any  other,  of 
the  same  number  of  sides,  and  of  equal  perimeter,  the  sides  are 

EQUAL. 

The  polygon  ABCDF  (Fig.  29.) 
cannot  be  a  maximum,  among  all 
polygons  of  the  same  number  of 
sides,  and  of  equal  perimeters,  un- 
less it  be  equilateral.  For  if  any 
two  of  the  sides,  as  CD  and  FD, 
are  unequal,  let  CH  and  FH  be 

lual,  and  their  sum  the  same  as 

lie  sum  of  CD  and  FD.      The 

')sceles  triangle  CHF  is  greater  than  the  scalene  triangle 
L'DF  (Art.  19.);  and  therefore  the  polygon  ABCIIF  is 
greater  than  the  polygon  ABCDF ;  so  that  the  latter  is  not 
a  maximum, 

PBOP08inoN*VL 

87.  A  REQULAB  POLraoN  has  a  greater  area  than  any 
other  polygon  of  equal  perimeter,  and  of  the  same  number  of 
sides. 


84  ISOPERIMETRY 

For,  by  tlie  preceding  article,. tlie  polygon  which  is  a  max- 
imum among  others  of  equal  perimeters,  and  the  same  num- 
ber of  sides,   is  equilateral,  and  by  Art.  84,  it  may  be  in- 
scribed in  a  circle.     But  if  a  poly- 
gon inscribed  in  a  circle  is  equilat- 
eral, as  ABDFGH,  it  is  also  equian- 
gular.   For  the  sides  of  the  polygon 
are  the  bases  of  so  many  isosceles 
triangles,  whose   common  vertex  is 
the  centre  C.     The  angles  at  these 
bases  are  all  equal ;  and  two  of  them, 

as  AHC  and  GHC,  are  equal  to  AHG  one  of  the  angles  of 
the  polygon,  ^he  polygon,  then,  being  equiangular,  as  well 
as  equilateral,  is  a  re^z^Zar  polygon.  (Art.  1.  Def.  2.) 

Thus  an  equilateral  triangle  has  a  greater  area,  than  any 
other  triangle  of  equal  perimeter.  And  a  square  has  a 
greater  area  than  any  other  four-sided  figure  of  equal  pe- 
rimeter; 

Cor.  A  regular  polygon  has  a  less  perimeter  than  any 
other  polygon  of  equal  area,  and  the  same  number  of 
sides. 

For  if,  with  a  given  perimeter,  the  regular  polygon  is 
greater  than  one  which  is  not  regular ;  it  is  evident  the  pe- 
rimeter of  the  former  must  be  diminished,  to  make  its  area 
equal  to  that  of  the  latter. 


Proposition  VII. 

88.  If  a  polggon  he  BESCRiBEB  about  a  circle,  the  areas 
of  the  two  figures  are  as  their  ^perimeters. 

Let  ST  be  one  of  the  sides  of  a  polygon,  either  regular  or 


ISOPERIMETRY. 


85 


T           1 

A            8 

A 

/.  \ 

V 

y 

not,  which  is  described  about  the  cir- 
cle LUR.  Join  OS  and  OT,  and 
to  the  point  of  contact  M  draw  the 
radius  OM,-«|»ich  will  be  perpen- 
dicular to  ST.  (Euc.  18,  3.)  The 
triangle  OST  is  equal  to  half  the 
liase  ST  multiplied  into  the  radius 
>M.  (Art.  8.)  And  if  lines  b« 
drawn,  in  the  same  manner,  from 

the  centre  of  the  circle,  to  the  extremities  of  the  sev- 
tral  sides  of  the  circumscribed  polygon,  each  of  the  trian- 
j;lcs  thus  formed  will  be  equal  to  half  its  base  multiplied 
into  the  radius  of  the  circle.  Therefore  t^e  area  of  the 
^vllole  polygon  is  equal  to  half  its  perimeter  multiplied  into 
the  radius  :  and  the  area  of  the  circle  is  equal  to  half  its  cir- 
cumference multiplied  into  the  radius.  (Art  30.)  So  that 
'  he  two  areas  aic  to  each  other  as  their  perimeters. 

Cor.  1.  If  diflferent  polygons  are  described  about  the 
>.ime  circle,  their  areas  are  to  each  other  as  their  perimeters. 
For  the  area  of  each  is  equal  to  half  its  perimeter,  multi- 
j)lied  into  the  radius  of  the  inscribed  circle. 

Cor.  2.  The  tangent  of  an  arc  is  always  greater  than  the 
arc  itself.  The  triangle  OMT  is  to  OMN,  as  MT  to  MN. 
But  OMT  is  greater  than  OMN,  because  the  former  includes 
the  latter.  Therefore,  the  tangent  MT  is  greater  than  the 
arc  MN. 


Proposition  VIII. 

89.  A  CIRCLE  has  a  greater  area  than  any  polygon  of  equal 
perimeter.  ^ 

If  a  circle  and  a  regular  polygon  have  the  same  centre, 
and  equal  perimeters ;  each  of  the  sides  of  the  polygon 
must  fall  partly  within  the  circle.    For  the  area  of  a  circum-» 


A  A 

1                \/o             1 

V 

A 

A^--^ 

r-^^^D 

86  ISOPERIMETRY. 

scribing  polygon  is  greater  tlian  the  area  of  the  circle,  as 
the  one  includes  the  other  :  and  therefore,  by  the  preceding 
article,  the  perimeter  of  the  former  is  greater  than  that  of 
the  latter. 

Let  AD  then  be  one  side  of  a 
regular  polygon,  whos*  perimeter 
is  equal  to  the  circumference  of 
the  circle  RLN.  As  this  falls 
partly  within  the  circle,  the  per- 
pendicular OP  is  less  than  the 
radius  OR.     But  the  area  of  the 

polygon  is  equal  to  half  its  pe-  " ~ 

rimeter  multiplied  into  this  per- 
pendicular (Art.  15.)  ;  and  the  area  of  the  circle  is  equal  to 
half  its  circumference  multiplied  into  the  radius.  (Art.  30.) 
The  circle  then  is  greater  than  the  given  regular  polygon ; 
and  therefore  greater  than  any  other  polygon  of  equal  pe- 
rimeter. (Art.  87.) 

Cor.  1.  A  circle  has  a  less  perimeter,  than  any  polygon  of 
equal  area. 

Cor.  2.  Among  regular  polygons  of  a  given  perimeter, 
that  which  has  the  greatest  number  of  sides,  has  also  the 
greatest  area.  For  the  greater  the  number  of  sides,  the 
more  nearly  does  the  perimeter  of  the  polygon  approach  to 
a  coincidence  with  the  circumference  of  a  circle. 


Proposition  IX. 

90.  A  right  prism  whose  bases  are  regular  polygons,  has 
a  less  surface  than  any  other  right  prism  of  the.  same  solidity , 
the  same  altitude,  and  the  same  number  of  sides. 

If  the  altitude  of  a  prism  is  given,  the  area  of  the  base  is 
as  the  soHdity  (Art.  43.) ;   and  if  the  number  of  sides  is 


isorKiiiM!:juv.  87 

perimeter  is  &  minimurn  vf hen  the  base  is  a 

i       ^  1.  (Art.  87.  Cor.)     But  the  lateral  surface  is 

s   the  perimeter.  (Art-  ^y.)^    Of  two  right  prisms,  then, 

vhich  have  the  same  altitude,  the  same  solidity,  and  the 

me  number  of  sides,  that  whose  bases  are  regular  polygons 

iias  the  least  lateral  surface,  while  the  areas  of  the  ends  are 

equal. 

Cor.  A  right  prism  whose  bases  are  regular  polygons  has 
a  greater  solidity,  than  any  other  right  prism  of  the  same 
surface,  the  same  altitude,  and  the  same  number  of  sides. 


PROPosmoN  X. 

91.  A  right  cruNDER  has  a  less  surface  than  any  right 
prism  of  the  same  altitude  and  solidity. 

For  if  the  prism  and  cylinder  have  the  same  altitude  and 
solidity,  the  areas  of  their  bases  are  equal.  (Art.  64.)  But 
the  perimeter  of  the  cylinder  is  less,  than  that  of  the  prism 
(Art.  89.  Cor.  1.) ;  and  therefore  its  lateral  surface  is  less, 
while  the  areas  of  the  ends  are  equal. 

Cor.  A  right  cylinder  has  a  greater  solidity,  than  any  right 
prism  of  the  same  altitude  and  surface. 

PuorosiTioN  XI. 

92.  A  CUBE  has  a  less  surface  titan  any  other  right  paral- 
lelepiped of  the  same  solidity. 

A  parallelepiped  is  a  prism,  any  one  of  whose  faces  may 
be  considered  a  base.  (Art.  41.  Def.  I  and  V.)  If  these  are 
not  all  squares,  let  one  which  is  not  a  square  be  taken  for  a 
base.  The  perimeter  of  this  may  be  diminished,  without 
altering  its  area  (Art.  87.  Cor.);  and  therefore  the  surface 


88  ISOPERIMETRY. 

of  tlie  solid  may  be  diminislied,  without  altering  its  altitude 
or  solidity.  (Art.  43,  47.)  The  same  may  be  proved  of 
each  of  the  other  faces  which  are  not  squares.  The  surface 
is  therefore  a  minimum,  when  all  the  faces  are  squares,  that 
is,  when  the  solid  is  a  cuhc. 

Cor.  A  cube  has  a  greater  solidity  than  any  other  right 
parallelopiped  of  the  same  surface. 

Proposition  XII. 

^93.  A  CUBE  has  a  greater  solidity  than  any  other  right  par- 
allelojriped,  the  sum  of  whose  length,  breadth  and  depth,  is  equal 
to  ike  sum  of  the  corresponding  dimensions  of  the  cube. 

The  solidity  is  equal  to  the  product  of  the  length,  breadth, 
and  depth.  If  the  length  and  breadth  are  unequal,  the 
solidity  may  be  increased,  without  altering  the  sum  of  the 
three  dimensions.  For  the  product  of  two  factors  whose 
sum  is  given,  is  the  greatest  when  the  factors  are  equal. 
(Euc.  27.  6.)  In  the  same  manner,  if*  the  breadth  and 
depth  are  unequal,  the  solidity  may  be  increased,  without 
altering  the  sum  of  the  three  dimensions.  Therefore,  the 
solid  cannot  be  a  maximum,  unless  its  length,  breadth,  and 
depth  are  equal. 

Proposition  XIII. 

94.     If  a     PRISM    BE     DESCRIBED     ABOUT     A    CYLINDER,     the 

cap>acities  of  the  two  solids  are  as  their  surfaces. 

The  capacities  of  the  solids  are  as  the  areas  of  their  bases, 
that  is,  as  the  perimeters  of  their  bases.  (Art.  88.)  But  the 
lateral  surfaces  are  also  as  the  perimeters  of  the  bases. 
Therefore  the  whole  surfaces  are  as  the  solidities. 

Cor.  The  capacities  of  different  prisms,  described  about 
the  same  right  cylinder,  are  to  each  other  as  their  surfaces. 


ISOrERIMETUY.  80 


Proposition  XIV. 

J...  J  right  cylinder  whose  height  is  equal  to  the 
DIAMETER  OF  ITS  BASE  hos  a  greater  solidity  t/iak  any  other 
right  cylinder  of  equal  surface. 

Let  C  be  a  rij^ht  cylinder  whose  lieight  is  equal  to  the  di- 
ameter of  its  base  ;  and  C  another  right  cylinder  having  the 
same  surface,  but  a  different  altitude.  If  a  square  prism  P 
be  described  about  the  former,  it  will  be  a  cube.  But  a 
square  prism  P'  described  about  the  latter  will  not  be  a  cube. 

Then  the  surfaces  of  C  and  P  are  as  their  bases  (Art.  47. 
and  88.) ;  which  arc  as  the  bases  of  C  and  P',  (Sup.  Euc. 
7,  1.);  80  that, 

surf  C  :  aurfV  : :  base  C  :  base  V : :  base  C  t  base  P' : :  surfC  \ 
mrfV. 

But  the  surface  of  C  is,  by  supposition,  equal  to  the  sur- 
face of  C  Therefore,  (Alg.  396.)  the  surface  of  P  is  equal 
to  the  surface  of  P'.     And  by  the  preceding  article, 

solid  P  :  solid  C  : :  surfP  :  sur/C  : :  siirfV  :  surfC  : :  solid 
V  :  solid  C. 

Cut  the  solidity  of  P  is  greater  than  that  of  P'.  (Art.  92. 
Cor.)    Therefore  the  solidity  of  C  is  greater  than  that  of  C. 

Schol.  A  right  cylinder  whose  height  is  equal  to  the  di- 
ameter of  its  base,  is  that  which  circumscribes  a  sjihere.  It 
is  also  called  Archimedes*  cylinder ;  as  he  discovered  the 
ratio  of  a  sphere  to  its  circumscribing  cylinder ;  and  these 
are  the  figures  which  were  put  upon  his  tomb. 

Cor.  Archimedes*  cylinder  has  a  less  surface^  than  any 
other  right  cylinder  of  the  same  capacity.  -Mt^'* 


90  ISOPERIMETRY. 


Proposition  XY. 


96.  If  a  SPHERE  BE  CIRCUMSCRIBED  hy  a  solid  hounded  hy 
'plane  surfaces  ;  the  capacities  of  the  two  solids  are  as  their 
surfaces. 

If  planes  be  supposed  to  be  drawn  from  the  centre  of  the 
sphere,  to  each  of  the  edges  of  the  circumscribing  soHd, 
they  will  divide  it  into  as  many  pyramids  as  the  solid  has 
faces.  The  base  of  each  pyramid  will  be  one  of  the  faces  ; 
and  the  height  will  be  the  radius  of  the  sphere.  The 
capacity  of  the  pyramid  will  be  equal,  therefore,  to  its  base 
multiplied  into  -^  of  the  radius  (Art.  48.) ;  and  the  capacity 
of  the  whole  circumscribing  solid,  must  be  equal  to  its  whole 
surface  multiplied  into  ^  of  the  radius.  But  the  capacity  of 
the  sphere  is  also  equal  to  its  surface  multiplied  into  ^  of  its 
radius.  (Art.  '70.) 

Cor.  The  capacities  of  different  solids  circumscribing  the 
same  sphere,  are  as  their  surfaces. 

Proposition  XVI. 

97.  A  SPHERE  has  a  greater  solidity  than  any  7'egularpoly- 
edron  of  equal  suyface. 

If  a  sphere  and  a  regular  polyedron  have  the  same  centre, 
and  equal  surfaces ;  each  of  the  faces  of  the  polyedron  must 
fall  partly  2vithin  the  sphere.  For  the  solidity  of  a  circum- 
scriUng  solid  is  greater  than  the  solidity  of  the  sphere,  as 
the  one  includes  the  other :  and  therefore,  by  the  preceding 
article,  the  surface  of  the  former  is  greater  than  that  of  the 
latter. 

But  if  the  faces  of  the  polyedron  fall  partly  within  the 
sphere,  their  perpendicular  distance  from  the  centre  must  be 
less  than  the  radius.     And  therefore,  if  the  surface  of  the 


180PERIMETRY.  91 

polyedron  be  only  equal  to  that  of  the  sphere,  its  solidity 
must  be  less.  For  the  solidity  of  the  polyedron  is  equal  to 
its  surface  multiplied  into  -J  of  the  distance  from  the  centre. 
(Art.  59.)  And  the  solidity  of  the  sphere  is  equal  to  its 
surface  multiplied  into  -^  of  the  radius. 

Cor.  A  sphere  has  a  less  surface  than  any  regular  poly- 
edron of  the  same  capacity. 


APPENDIX 


GAUGING     OF     CASKS. 

Art.  119.  Gauging  is  a  practical  art,  which  does  not  ad- 
mit of  being  treated  in  a  very  scientific  manner.  Casks  are 
not  commonly  constructed  in  exact  conformity  with  any  reg- 
ular mathematical  figure.  By  most  writers  on  the  subject, 
however,  they  are  considered  r  as  nearly  coinciding  with  one 
of  the  following  forms : 

1.  ?  ^,        .-,,,/.  (  of  a  spheroid, 

2.  1  ^^^  ^^^^^"  ^^^'*^^  i  of  a  parabolic  spindle. 
3-  I  rri..  ..„„!  ^.„..,„..  S  ^f  a  paraboloid. 


.'  f  The  equal  frustums  ]    . 

4.  )  ^  (  of  a  cone. 


The  second  of  these  varieties  agrees  more  nearly  than  any 
of  the  others,  with  the  forms  of  casks,  as  they  are  com- 
monly made.  The  first  is  too  much  curved,  the  third  too 
little,  and  the  fourth  not  at  all,  from  the  head  to  the  bung. 

120.  Rules  have  already  been  given,  for  finding  the  capa- 
city of  each  of  the  four  varieties  of  casks.  (Arts.  68,  110, 
112,  118.)  As  the  dimensions  are  taken  in  mcAes,  these  rules 
will  give  the  contents  in  cubic  inches.  To  abridge  the  com- 
putation, and  adapt  it  to  the  particular  measures  used  in 
gauging,  the  factor  .'7854  is  divided  by  282  or  231 ;  and 
the  quotient  is  used  instead  of  .7854,  for  finding  the  capa- 
city in  ale  gallons  or  wine  gallons. 


GAUGING.  9S 

Now ^^rr^  =.002785,  or  .0028  nearly  ; 

And  -l^^.OOSi 
231 

If  then  .0028  and  .0034  be  substituted  for  .'7854,  in  the 
rules  referred  to  above;  the  contents  of  the  cask  will  be 
given  in  ale  gallons  and  wine  gallons.  These  numbers  are 
to  eacli  other  nearly  as  9  to  11. 

Problem  I. 

To  calculate  the  contents  of  a  cask,  in  the  form  of  a  middle 
frustum  of  a  spheroid. 

121.  Add  together  the  square  of  the  head  diameter,  and 
twice  the  square  of  the  bung  diameter  :  multiply  the  sum  by 
^  of  the  length,  and  the  product  by  .0028  for  ale  gallons,  or 
by  .0034  for  wine  gallons. 

If  D  and  fl?=the  two  diameters,  and  /=the  length  ; 

The  capacity  in  inches=(2D'+c?')Xi/X.'7854.  (Art.  110.) 

And  by  substituting  .0028  or  .0034  for  .7854,  we  have 
the  capacity  in  ale  gallons  or  wine  gallons. 

Ex.  What  is  the  capacity  of  a  cask  of  the  first  form, 
whose  length  is  30  inches,  its  head  diameter  18,  and  its  bung 
diameter  24  ? 

Ans.  41.3  ale  gallons,  or  50.2  wine  gallons. 

Problem  II. 

To  calculate  t/u:  contents  of  a  cask,  in  the  form  of  tlie  mid- 
dle frustum  of  a  parabouc  spixdle. 

122.  Add  together  the  square  of  the  head  diameter,  and 
twice  the  square  of  the  bung  diameter,  and  from  the  sum 


94  GAUGING. 

subtract  |  of  the  square  of  the  difference  of  the  diaraeters ; 
multiply  the  remainder  by  -^  of  the  length,  and  the  product 
by  .0028  for  ale  gallons,  or  .0034  for  wine  gallons. 

The  capacity  in  inches  ={2J)^+cP—l  (D—dy)x¥X 
.Y854.   (Art.  118.) 

Ex.  What  is  the  capacity  of  a  cask  of  the  second  form, 
whose  length  is  30  inches,  its  head  diameter  18,  and  its 
bung  diameter  24  ? 

Ans.  40.9  ale  gallons,  or  49.7  wine  gallons. 

Problem  III. 

To  calculate  the  contents  of  a  cask,  in  the  form  of  two  equal 
frustums  of  a  paraboloid. 

123.  Add  together  the  square  of  the  head  diameter,  and 
the  square  of  the  bung  diameter  ;  multiply  the  sum  by  half 
the  length,  and  the  product  by  .0028  for  ale  gallons,  or 
.0034  for  wine  gallons. 

The  capacity  in  inches  =(D'+d')xilX.'7854.  (Art.  112 
Cor.) 

Ex.  What  is  the  capacity  of  a  cask  of  the  third  form, 
■whose  dimensions  are,  as  before,  30,  18,  and  24  ? 

Ans.  37.8  ale  gallons,  or  45.9  wine  gallons. 

Problem  IV. 

To  calculate  the  contents  of  a  cash,  in  the  form  of  tivo  equal 
frustums  of  a  coine. 

124.  Add  together  the  square  of  the  head  diameter,  the 
square  of  the  bung  diameter,  and  the  product  of  the  two 
diameters  ;  multiply  the  sum  by  \  of  the  length,   and  the 
product  by.0028  for  ale  gallons,  or  .0034  for  wine  gallons. 
The  capacity  in  inches=(D-+c^'+Dc?)x-J^X.7854.  (Art.  68.) 


GAUGINO.  95 

Ex.  What  is  the  capacity  of  a  cask  of  the  fourth  form, 
whoso  length  is  30,  and  its  diameters  18  and  24  ? 

Ans.  37.3  .ale  gallons,  or  45.3  wine  gallons. 

12o.  The  precedini^  rules,  though  correct  in  theory,  are 
not  very  well  adapted  to  practice,  as  they  suppose  the  form 
of  the  cask  to  be  Jcnoion^  The  two  following  rules,  taken 
from  Hut  ton's  Mensuration,  may  be  used  for  casks  of  the 
usual  forms.  For  the  first,  three  dimensions  are  required, 
the  length,  the  head  diameter,  and  the  bung  diameter.  It 
is  evident  tliat  no  allowance  is  made  by  this,  for  different 
degrees  of  curvature  from  the  head  to  the  bung.  If  the 
cask  is  more  or  less  curved  than  usual,  the  following  rule  is 
to  be  preferred,  for  which /owr  dimensions  are  required,  the 
head  and  bung  diameters,  and  a  third  diameter  taken  in  the 
middle  between  the  bung  and  the  head.  For  the  demon- 
stration of  these  rules,  see  Hutton's  Mensuration,  Part  V. 
Sec.  2.  Ch.  5  and  7. 

Problem  V. 

To  calculate  the  contents  of  any  common  cask,  from  three 
dimcnsums. 

126.  Add  together 

25  times  the  square  of  the  head  diameter, 
39  times  the  square  of  the  bung  diameter,  and 
20  times  the  product  of  the  two  diameters ; 
Multiply  the  sum  by  the  length,  divide  the  product  by  90, 
and  multiply  the  quotient  by  .0028  for  ale  gallons,  or  .0034 
for  wine  gallons. 

The  capacity  in  inches=(39  D'-f25cr'4-26D(/)x  -  X. 7854. 

Ex.  Wliat  is  the  capacity  of  a  cjisk  whose  length  is  30 
inches,  the  head  diameter  18,  and  the  bung  diameter  24? 
Ans,  39  ale  gallons,  or  47^  wine  gallons. 


96  GAUGING. 


Problem  VI. 

2h  calculate  the  contents  of  a  cask  from  four  dimensions,  the 
length,  the  head  and  hung  diameters,  and  a  diameter  taken 
in  the  middle  between  the  head  and  the  hung. 

127.  Add  together  the  squares  of  the  head  diameter,  of 
the  bung  diameter,  and  of  double  the  middle  diameter ; 
multiply  the  sum  by  -^  of  the  length,  and  the  product  by 
.0028  for  ale  gallons,  or  .0034  for  wine  gallons. 

If  D=the  bung  diameter,  c:?=the  head  diameter,  m=the 
middle  diameter,  and  Z=the  length  ; 

The  capacity  in  inches=(D'-|-c?'+2m^)Xi?X.'7854. 

Ex.  What  is  the  capacity  of  a  cask,  whose  length  is  30 
inches,  the  head  diameter  18,  the  bung  diameter  24,  and  the 
middle  diameter  22^  ? 

Ans.  41  ale  gallons,  or  49|  wine  gallons. 

128.  In  making  the  calculations  in  gauging,  according  to 
the  preceding  rules,  the  multiplications  and  divisions  are  fre- 
quently performed  by  means  of  a  Sliding  Rule,  on  which 
are  placed  a  number  of  logarithmic  lines,  similar  to  those  on 
Gunter's  Scale.  See  Trigonometry,  Sec.  VI.,  and  Note  C. 
p.  149. 

Another  instniment  commonly  used  in  gauging  is  the 
Diagonal  Rod.  By  this,  the  capacity  of  a  cask  is  very  ex- 
peditiously found,  from  a  single  dimension,  the  distance  from 
the  bung  to  the  intersection  of  the  opposite  stave  with  the 
head ;  but  this  process  is  not  considered  sufficiently  accurate 
for  casks  of  a  capacity  exceeding  40  gallons.  The  measure 
is  taken  by  extending  the  rod  through  the  cask,  from  the 
bung  to  the  most  distant  part  of  the  head.  The  number  of 
gallons  corresponding  to  the  length  of  the  line  thus  found,  is 
marked  on  the  rod.     The  logarithmic  lines  on  the  gauging 


GAUGING.  97 

rod  are  to  be  used  in  the  same   manner,  as  on  the  sliding 
rule. 

ULLAGE    OF    CASKS. 

129.  When  a  cask  is  partly  filled,  the  whole  capacity  is 
divided,  by  the  surface  of  the  liquor,  into  two  portions ;  the 
least  of  which,  whether  full  or  empty,  is  called  the  ullage. 
In  finding  the  ullage,  t]«e  cask  is  supposed  to  be  in  one  of 
two  positions  ;  either  standinr/,  with  its  axis  perpendicular  to 
the  horizon ;  oj-  lyi7i(/,  with  its  axis  parallel  to  the  horizon. 
The  rules  for  ullage  which  are  exact,  particularly  those  for 
lying  casks,  are  too  complicated  for  common  use.  The  fol- 
lowing are  considered  as  sufficiently  near  approximations. 
See  Hutton's  Mensuration. 


Problem  VII. 

To  calculate  the  ullage  of  a  standing  cask. 

130.  Add  together  the  squares  of  the  diameter  at  the  sur- 
face of  the  liquor,  of  the  diameter  of  the  nearest  end,  and 
of  double  the  diameter  in  the  middle  between  the  other  two ; 
multiply  the  sum  by  \  of  the  distance  between  the  surface 
and  the  nearest  end,  and  the  product  by  .0028  for  ale  gal- 
lons, or  .0034  for  wine  gallons. 

If  D=the  diameter  of  the  surface  of  the  liquor, 
rf=the  diameter  of  the  nearest  end, 
m=the  middle  diameter,  and 

/=the  distance  between  the  surface  and  the  nearest  end ; 
The  ullage  in  inches=(D'+rf'+2m^)Xi/X-V8o4. 

Ex.  If  the  diameter  at  the  surface  of  the  liquor,  in  a  stand- 
ing cask,  be  32  inches,  the  diameter  of  the  nearest  end  24, 
the  middle  diameter  29,  and  the  distance  between  the  sur- 

5 


98  GAUGING. 

face  of  the  liquor  and  the  nearest  end  1 2  ;  what  is  the  ul- 
lage? Ans.  2li  ale  gallons,  or  33^  wine  gallons. 

Problem  VIII. 
To  calculate  the  ullage  of  a  lying  cask. 

131.  Divide  the  distance  from  the  buna:  to  the  surface  of 
the  liquor,  by  the  whole  bung  diameter,  find  the  quotient  in 
the  column  of  heights  or  versed  sines  in  a  table  of  circular 
segments,  take  out  the  corresponding  segment,  and  multiply 
it  by  the  whole  capacity  of  the  cask,  and  the  product  by  1-^ 
for  the  part  which  is  empty. 

If  the  cask  be  not  half  full,  divide  the  depth  of  the  liquor 
by  the  whole  bung  diameter,  take  out  the  segment,  multiply, 
&c.,  for  the  contents  of  the  part  which  is  full. 

Ex.  If  the  whole  capacity  of  a  lying  cask  be  41  ale  gal- 
lons, or  49-f  wine  gallons,  the  bung  diameter  24  inches  and 
the  distance  from  the  bung  to  the  surface  of  the  liquor  6 
inches  ;  what  is  the  ullage  ? 

Ans.  Vf-  ale  gallons,  or  9^  Avine  gallons. 


NOTES 


Note  A.  p.  39.    , 

The  term  solidity  i»  used  here  in  the  customary  sense,  to 
express  the  magnitude  of  any  geometrical  quantity  of  three 
dimensions,  length,  breadth,  and  thickness ;  whether  it  be  a 
solid  body,  or  a  fluid,  or  even  a  portion  of  empty  space.  This 
use  of  the  word,  however,  is  not  altogether  free  from  objec- 
tion. The  same  term  is  applied  to  one  of  the  general  prop- 
erties of  matter  ;  and  also  to  that  peculiar  quality  by  which 
certain  substances  are  distinguished  from  fluids.  There 
seems  to  be  an  impropriety  in  speaking  of  the  solidity  of  a 
body  of  water^  or  of  a  vessel  which  is  ejnpty.  Some  writers 
have  therefore  substituted  the  word  volume  for  solidity.  But 
the  latter  term,  if  it  be  properly  defined,  may  be  retained 
without  danger  of  leading  to  mistake. 

Note  B.  p.  76. 

The  following  simple  rule  for  the  solidity  of  round  timber, 
or  of  any  cylinder,  is  nearly  exact : 

Multiply  the  length  into  twice  the  square  of  ^  of  the  circum,' 
fereiice. 

If  C=the  circumference  of  a  cylinder; 

The  area  of  the  base=-^=— ^^—  But  2(^\  =-^1- 
4^     12.566  \5/       12.5 

It  is  common  to  mea.siire  heton  timber,  by  multiplying  the 
length  into  the  square  of  the  quarter-ffirt.     This  gives  ex- 


100  NOTES. 

actly  the  solidity  of  a  parallelopiped,  if  tLc  ends  are  squares. 
But  if  the  ends  are  parallelograms,  the  area  of  each  is  less 
than  the  square  of  the  quarter-girt.  (Euc.  27.  6.) 

Timber  which  is  taiKring  may  be  exactly  measured  by  the 
rule  for  the  frustum  of  a  pyramid  or  cone  (Art.  50,  68.) ; 
or,  if  the  ends  are  not  similar  figures,  by  the  rule  for  a  pris- 
moid.  (Art.  55.)  But  for  common  purposes,  it  will  be  suf- 
ficient to  multiply  the  length  by  the  area  of  a  section  in  the 
middle  between  the  two  ends. 


^ 

4 

NATURAL 

TAXCKN'TS. 

M 

0 

44  Degrees. 

M 

60 

31 

44  Degrees. 

M 

29 

N.Tan. 
1)0509 

N.Cot. 
1.03553 

N.  Til  11. 

N.  Cot. 

98327 

1.01702 

1 

90025 

1.03493 

59 

.32 

98384 

1.01642 

28 

2 

90081 

1.03433 

58 

33 

98441 

1.01583 

27 

3 

90738 

1.03372 

57 

34 

98499 

1.01524 

26 

4 

9671)4 

1.03312 

56 

35 

98556 

1.01405 

25 

5 

96850 

1.03252 

55 

36 

98613 

1.01406 

24 

6 

90907 

1.03192 

54 

37 

98071 

1.01347 

23 

7 

90963 

1.03132 

53 

38 

98728 

1.01288 

22 

8 

97020 

1.03072 

52 

39 

98780 

1 .01229 

21 

9 

97076 

1.03012 

51 

40 

98843 

1.01170 

20 

10 

97133 

1.02952 

50 

41 

98901 

1.01112 

19 

11 

97189 

1.02892 

49 

42 

98958 

1.01053 

18 

11' 

97246 

1.05832 

48 

43 

99016 

1.00994 

17 

•^■»02 

1.02772 

47 

44 

99073 

1.00935 

16 

14 

97359 

1.02713 

40 

45 

99131 

1.00876 

15 

15 

97416 

1.02053 

45 

46 

99189 

1.00818 

14 

16 

97472 

1.02.193 

44 

47 

99247 

1.00759 

13 

17 

97529 

1.02533 

43 

48 

99304 

1.00701 

12 

18 

97586 

1.02474 

42 

49 

99302 

1.00042 

11 

...  1 

«.        19 

97043. 

1.02414 

41 

50 

99420 

1.00583 

10 

"■      20 

•97700- 

1.02355 

40 

51 

99478 

1.00525 

9' 

21 

97756 

1.02295 

39 

52 1 

W^^Q 

8 

•        22 
23 

a:8i3 

1.02236 
^1.02176 

38 
37 

5? 

99594  ,' 
'  99b'52 

'l-oosl 

7       . 
6 

24 

9792'? 

1.02117 

36 

55 

99710 

l! 00291 

5 

25 

97984 

1.02057 

35 

56 

99708 

1.00233 

4 

26 

98041 

1.01998 

34 

57 

99326 

1.00175 

3 

27 

98098 

1.01939 

33 

58 

99884 

1.00116 

2 

28 

98155 

1.01879 

32 

59 

99942 

1.00058 

1 

29 

98213 

1.0L820 

31 

CO 

10000 

1.00000 

0 

-       ?^. 

9§?70 

1.01^ 

30 

M     38 

M 

N.  Cot. 

N.  Tan.  • 

M 

HT 

N.Cot. 

N.  Tan. 

45  Degrees. 

^5D^ 

grees. 

-■ 

■ 

i 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 
LOAN  DEPT. 

This  book  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 

Renewed  books  are  subject  to  immediate  recall. 


^:>\  mcnt.  S"?    kk| 


REC'D  LD 

'    NQV  13  1957 


I7MaroCA, 


n^.-O  LD 


260ct59FC 


MAii-^es--4m 


REC'D  LD 


OCT  2  6  1959 


I^CgTWTKHi 


IJC 


,D 


A. 


NOV  1  .>  1959 


^-Rl 


\0^- 


twtVJ^ 


^ 


^ 


4' 


DEC  4  ■  19C9 


LD  21A-50m-8,'57 
(C848l8l0)476B 


4 


General  Library 

University  of  California 

Berkeley 


